Permutations and Combinations

To solve this problem, we can use a classic combinatorics method known as "stars and bars" (or "balls and bins"), which is a way to solve problems involving distributing identical items into distinct groups with certain restrictions.

First, since each child must get at least 2 apples, let's give 2 apples to each child right away. That accounts for 6 apples (2 apples for each of the 3 children). Now, we have 21 - 6 = 15 apples left to distribute freely among the three children.

The "stars and bars" technique involves representing the apples as stars (*) and the divisions between children as bars (|). For example, if we had 5 apples to distribute among three children, one possible distribution could be represented as **|*|**. This means the first child gets 2 apples, the second child gets 1 apple, and the third child gets 2 apples.

In our case, we need to distribute 15 apples (stars) among the three children with 2 bars to create the partitions. We arrange 15 stars and 2 bars in a row, where the arrangement of stars and bars corresponds to a distribution of the apples.

The total number of objects we're arranging is 15 apples + 2 bars = 17 objects. We need to choose 2 positions out of these 17 to place the bars. The remaining positions will be occupied by the stars (apples).

The number of ways to choose 2 positions out of 17 for the bars is given by the binomial coefficient:

Thus, there are 136 ways to distribute the 21 identical apples among three children such that each child gets at least 2 apples. The correct answer is Option B: 136.

$\begin{aligned} & { }^6 \mathrm{C}_{\mathrm{m}}+2\left({ }^6 \mathrm{C}_{\mathrm{m}+1}\right)+{ }^6 \mathrm{C}_{\mathrm{m}+2}>{ }^8 \mathrm{C}_3 \\ & { }^7 \mathrm{C}_{\mathrm{m}+1}+{ }^7 \mathrm{C}_{\mathrm{m}+2}>{ }^8 \mathrm{C}_3 \\ & { }^8 \mathrm{C}_{\mathrm{m}+2}>{ }^8 \mathrm{C}_3 \\ & \therefore \mathrm{m}=2 \\ & \text { And }{ }^{\mathrm{n}-1} \mathrm{P}_3:{ }^n \mathrm{P}_4=1: 8 \\ & \frac{(\mathrm{n}-1)(\mathrm{n}-2)(\mathrm{n}-3)}{\mathrm{n}(\mathrm{n}-1)(\mathrm{n}-2)(\mathrm{n}-3)}=\frac{1}{8} \\ & \therefore \mathrm{n}=8 \\ & \therefore{ }^{\mathrm{n}} \mathrm{P}_{\mathrm{m}+1}+{ }^{\mathrm{n}+1} \mathrm{C}_{\mathrm{m}}={ }^8 \mathrm{P}_3+{ }^9 \mathrm{C}_2 \\ & =8 \times 7 \times 6+\frac{9 \times 8}{2} \\ & =372 \end{aligned}$

$3 \text { Shelf empty: }(8,0,0,0) \rightarrow 1 \text { way }$

$\left.2 \text { shelf empty: } \begin{array}{c} (7,1,0,0) \\ (6,2,0,0) \\ (5,3,0,0) \\ (4,4,0,0) \end{array}\right] \rightarrow 4 \text { ways }$

$\left.1 \text { shelf empty: } \begin{array}{cc} (6,1,1,0) & (3,3,2,0) \\ (4,2,1,0) & (4,2,2,0) \\ (4,3,0) & \end{array}\right] \rightarrow 5 \text { ways }$

$\left.0 \text { Shelf empty : } \begin{array}{ll} (1,2,3,2) & (5,1,1,1) \\ (2,2,2,2) \\ (3,3,2,1,1) \\ (4,2,1,1) & \end{array}\right] \rightarrow 5 \text { ways }$

Total $=15$ ways

$\begin{aligned} & \alpha=\frac{(4 !) !}{(4 !)^{3 !}}, \beta=\frac{(5 !) !}{(5 !)^{4 !}} \\ & \alpha=\frac{(24) !}{(4 !)^6}, \beta=\frac{(120) !}{(5 !)^{24}} \end{aligned}$

Let 24 distinct objects are divided into 6 groups of 4 objects in each group.

No. of ways of formation of group $=\frac{24 !}{(4 !)^6 .6 !} \in \mathrm{N}$

Similarly,

Let 120 distinct objects are divided into 24 groups of 5 objects in each group.

No. of ways of formation of groups

$=\frac{(120) !}{(5 !)^{24} \cdot 24 !} \in \mathrm{N}$

We need to find the sum of all the 4-digit numbers formed using the digits $\{0,3,6,9\}$ without repetition. Note that since the number is 4-digit, the thousands place cannot be $0$.

Step 1. Count the Valid Numbers

Thousands Place: Can be chosen from $\{3,6,9\}$ — $3$ possibilities.

Other Places: After choosing the thousands digit, the remaining three digits can be arranged in $3!$ ways (i.e. $6$ ways).

Thus, the total number of valid numbers is:

$ 3 \times 6 = 18. $

Step 2. Sum by Place Value Contribution

We calculate the contribution to the total sum from each digit place (thousands, hundreds, tens, and units).

(a) Thousands Place

Allowed digits: $3, 6, 9$.

Frequency: Each digit (3, 6, 9) appears in the thousands place $6$ times.

Contribution:

$ \text{Sum}_{\text{thousands}} = (3+6+9) \times 6 \times 1000 = 18 \times 6 \times 1000 = 108000. $

(b) Hundreds Place

For the hundreds place, consider the following:

Case 1: When thousands digit is $3$, the remaining digits are $\{0,6,9\}$. In these $6$ numbers, each of $0$, $6$, and $9$ appears exactly $2$ times in the hundreds place.

Case 2: When thousands digit is $6$, the remaining digits are $\{0,3,9\}$. Each appears $2$ times.

Case 3: When thousands digit is $9$, the remaining digits are $\{0,3,6\}$. Each appears $2$ times.

Thus, the total frequency in the hundreds place is:

$0:$ appears $2+2+2 = 6$ times.

$3:$ appears in cases 2 and 3: $2+2 = 4$ times.

$6:$ appears in cases 1 and 3: $2+2 = 4$ times.

$9:$ appears in cases 1 and 2: $2+2 = 4$ times.

Contribution:

$ \text{Sum}_{\text{hundreds}} = (0\cdot6 + 3\cdot4 + 6\cdot4 + 9\cdot4) \times 100. $

First calculate the inner sum:

$ 3\cdot4 + 6\cdot4 + 9\cdot4 = 4(3+6+9) = 4 \times 18 = 72. $

So,

$ \text{Sum}_{\text{hundreds}} = 72 \times 100 = 7200. $

(c) Tens Place

By symmetry (since after the thousands digit is fixed, the remaining three digits are uniformly permuted among the hundreds, tens, and units places), the distribution for the tens place is identical to that for the hundreds place.

Contribution:

$ \text{Sum}_{\text{tens}} = 72 \times 10 = 720. $

(d) Units Place

Again, by similar reasoning, the units place has the same digit frequency as the hundreds and tens places.

Contribution:

$ \text{Sum}_{\text{units}} = 72 \times 1 = 72. $

Step 3. Total Sum

Add the contributions from all four places:

$ \text{Total Sum} = 108000 + 7200 + 720 + 72. $

Calculate step-by-step:

$ 108000 + 7200 = 115200, $

$ 115200 + 720 = 115920, $

$ 115920 + 72 = 115992. $

Final Answer

The sum of all the 4-digit numbers formed is 115992.

Thus, the correct option is:

Option B – 115992.

Every things has two choices to go in to lst box and there are 6 such things there are two cases in which all the things go into lst or 2 nd box.

$\because$ Total number of ways $=2^6-2=62$

To find the number of ways in which the number 831600 can be divided into two factors that are relatively prime, we start by determining the prime factorization of 831600:

$ M = 831600 = 2^4 \times 3^3 \times 5^2 \times 7 \times 11 $

The number of different prime factors is 5: $2, 3, 5, 7, 11$.

When dividing a number into two relatively prime factors, each prime factor must entirely go to one factor or the other. The general formula to calculate the number of ways to partition these prime factors into two groups is given by:

$ 2^{n-1} $

where $n$ represents the number of distinct prime factors. Thus, for 831600:

$ 2^{5-1} = 2^4 = 16 $

Therefore, there are 16 different ways to split 831600 into two factors that are relatively prime.

Given, word Probability

Number of letter in probability

P,R,O,A,L,T,Y,BB,II

(i) 4 letter are selected alone different

i.e. $ { }^{9} C_{4}=\frac{9!}{4!5!}=126 $

(ii) Two letter or alike other are different

$ 2 C_{1} \times 8 C_{2}=\frac{2 \times 8!}{2!6!}=56 $

(iii) Two letter are alike

$ 2 C_{2}=1 $

Total number of sample space

$ =126+56+1=183 $

Favorable case in which at least one letter is repeated

$ \text { i.e., } \quad 56+1=57 $

$ \therefore \quad \text { Reqmend probability }=\frac{57}{183}=\frac{19}{61} $

To find the number of ways to arrange all the letters of the word 'COMBINATIONS' around a circle such that no two vowels are together, let's break it down step-by-step.

Identify Vowels and Consonants:

Vowels: O, O, I, I, A (total of 5 vowels)

Consonants: C, M, B, N, T, N, S (total of 7 consonants)

Arrangement of Consonants:

Since the arrangement is in a circle, we arrange the consonants in $(7-1)! = 6!$ ways because in circular permutations, we fix one position to break the circle and linearly order the rest.

Interleaving Vowels:

Arranging the consonants creates 7 gaps (before the first consonant, between each pair, and after the last consonant) to place the vowels. We need to select 5 gaps for the vowels, ensuring no two vowels are adjacent, which is crucial since we want them not together.

Arrangement of Vowels:

Arrange the vowels O, O, I, I, A in these gaps.

The arrangement of the vowels is affected by repetition. The formula for arranging 'n' items where there are repetitions is given by $\frac{n!}{k1! \cdot k2! \cdot \ldots \cdot kr!}$, where $k_i$ represents the number of times each repeated letter occurs.

The necessary calculation for the vowels: $\frac{5!}{2! \cdot 2!}$ because O and I each repeat twice.

Total Arrangements:

Combine the arrangements of consonants and vowels as follows:

$ \text{Total ways} = 6! \times \frac{5!}{2! \times 2!} $

By further expanding this calculation based on the repetitions:

$ \frac{7! \times 6!}{(2!)^4} $

This gives the total number of arrangements of the letters in 'COMBINATIONS' around a circle, ensuring that no two vowels are adjacent.

Explanation

To find the difference between the number of odd and even numbers formed by the digits 3, 5, 6, 7, and 8, where the numbers are greater than 6000 and less than 10000, we consider the following process:

Odd Numbers:

Set I: When the first digit (thousands place) is 6, 7, or 8, and the last digit (units place) is 3.

Choices:

3 choices for the first digit (6, 7, 8).

3 choices for the second digit.

2 choices for the third digit.

1 choice for the last digit (3).

Combinations:

$ 3 \times 3 \times 2 \times 1 = 18 $

Set II: Similar to Set I, but the last digit (units place) is 5.

Choices:

3 choices for the first digit (6, 7, 8).

3 choices for the second digit.

2 choices for the third digit.

1 choice for the last digit (5).

Combinations:

$ 3 \times 3 \times 2 \times 1 = 18 $

Set III: The first digit (thousands place) is 6 or 8, and the last digit is 7.

Choices:

2 choices for the first digit (6, 8).

3 choices for the second digit.

2 choices for the third digit.

1 choice for the last digit (7).

Combinations:

$ 2 \times 3 \times 2 \times 1 = 12 $

Total Odd Numbers:

$ 18 + 18 + 12 = 48 $

Even Numbers:

Set I: The first digit (thousands place) is 7 or 8, and the last digit (units place) is 6.

Choices:

2 choices for the first digit (7, 8).

3 choices for the second digit.

2 choices for the third digit.

1 choice for the last digit (6).

Combinations:

$ 2 \times 3 \times 2 \times 1 = 12 $

Set II: The first digit (thousands place) is 6 or 7, and the last digit (units place) is 8.

Choices:

2 choices for the first digit (6, 7).

3 choices for the second digit.

2 choices for the third digit.

1 choice for the last digit (8).

Combinations:

$ 2 \times 3 \times 2 \times 1 = 12 $

Total Even Numbers:

$ 12 + 12 = 24 $

Calculating the Difference:

The required difference between the number of odd and even numbers is:

$ 48 - 24 = 24 $

This result can be expressed as $ {}^4 P_3 $.

There are 4 possibilities for inviting the relatives to the party:

3 ladies from the husband's side and 3 gentlemen from the wife's side:

Total number of ways:

$ = { }^{4} C_{3} \times { }^{4} C_{3} = 16 $

3 gentlemen from the husband's side and 3 ladies from the wife's side:

Total number of ways:

$ = { }^{3} C_{3} \times { }^{3} C_{3} = 1 $

2 ladies and 1 gentleman from the husband's side and 2 gentlemen and 1 lady from the wife's side:

Total number of ways:

$ = \left( { }^{4} C_{2} \times { }^{3} C_{1} \right) \times \left( { }^{3} C_{1} \times { }^{4} C_{2} \right) = 324 $

1 lady and 2 gentlemen from the husband's side, and 2 ladies and 1 gentleman from the wife's side:

Total number of ways:

$ = \left( { }^{4} C_{1} \times { }^{3} C_{2} \right) \times \left( { }^{3} C_{2} \times { }^{4} C_{1} \right) = 144 $

The total number of ways to invite 3 of the man's relatives and 3 of the wife's relatives is:

$ = 16 + 1 + 324 + 144 = 485 $

Alphabets present in the word 'COLLEGE' are C, O, L, E, G

Alphabets in ascending order are

C, E, G, L, O

Number of words starts with $ \mathrm{CE}=\frac{5!}{2!} $

Number of words starts with CG $ =\frac{5!}{2!2!} $

Number of words starts with $ \mathrm{CL}=\frac{5!}{2!} $

Number of words starts with $ \mathrm{COE}=\frac{4!}{2!} $

Number of words starts with COG $ =\frac{4!}{2!2!} $

Number of words starts with COLE $ =3 $ !

Number of words starts with COLG $ =\frac{3!}{2!} $

Number of words starts with

COLLEE $ =1 $ !

Next words in COLLEGE

$ \therefore $ Required number of ways

$ \begin{array}{l} =\frac{5!}{2!}+\frac{5!}{2!2!}+\frac{5!}{2!}+\frac{4!}{2!}+\frac{4!}{2!2!}+3!+\frac{3!}{2!}+1+1 \\\\ =60+30+60+12+6+6+3+2 \\\\ =179 \end{array} $

Case I Let one digit be 1 or 7 , then sum of the other two digits be $ 2,5,8 $, 11 or 14 or $ 3 x+2 $ Therefore, the 2 digits can be $ (3,5),(5,9) $.

$ \therefore $ The possible number are 8 .

Case II Let one digit be 3 or 9 the sum of other two digits can be multiple of 3 . The other 2 digit can be $ (1,5) $ or $ (7,5) $ $ \therefore $ The possible numbers are 8 .

Case III Let one digit can be 5 , then sum of other digit can be $ 1,4,7,10,13 $ or 16.

$ \therefore $ The other 2 digit numbers can be $ (1,3),(1,9),(3,7),(7,6) $

$ \therefore $ The possible number can be 8 in $ ^{-} $ number.

Hence, the required number are

$ 8+8+8=24 $

We have, part A has 7 questions, Part B has 5 questions and part $ C $ has 3 questions.

Case I 4 questions from part $ A $, then 3 questions from part $ B $ and $ C $

Number of ways

$ ={ }^{7} C_{4} \times\left[{ }^{5} C_{3}+{ }^{5} C_{2} \times{ }^{3} C_{1}+{ }^{5} C_{1} \times{ }^{3} C_{2}\right] $

$ =\frac{7 \times 6 \times 5}{1 \times 2 \times 3}[10+30+15] $

$ =35 \times 55=1925 $

Case II 3 questions from Part $ A $, then 4 questions from part $ B $ and $ C $.

$ \begin{array}{l} \therefore \text { Number of ways } \\\\ ={ }^{7} C_{3} \times\left[{ }^{5} C_{3} \times{ }^{3} C_{1}+{ }^{5} C_{2} \times{ }^{3} C_{2}\right] \\\\ =35 \times[30+30]=35 \times 60=2100 \end{array} $

Case III 2 questions from part $ A $, then 5 question from part $ A $ and $ B $

Number of ways $ ={ }^{7} C_{2}\left[{ }^{5} C_{3} \times{ }^{3} C_{2}\right] $

$ =21 \times 10 \times 3=630 $

$ \therefore $ Required number of ways

$ =1925+2100+630=4655 $

The numbers will be divisible by 6 if and only if the unit digit of numbers is an even number and the sum of digits is divisible by 3 .

Collection of 4 -digits whose sum is divisible by 3 .

(1236) $\rightarrow \ldots \ldots \ldots . .2$ or $6 \rightarrow 3 \times 2 \times 1 \times 2=12$

$(1245) \rightarrow \ldots \ldots \ldots 2$ or $4 \rightarrow 3 \times 2 \times 1 \times 2=12$

$(1356) \rightarrow \ldots \ldots \ldots 6 \rightarrow 3 \times 2 \times 1 \times 1=6$

$(2346) \rightarrow \ldots \ldots \ldots .2$ or 4 or $6 \rightarrow$

$ 3 \times 2 \times 1 \times 3=18 $

(3456) $\rightarrow \ldots \ldots . .4$ or $6 \rightarrow 3 \times 2 \times 1 \times 2=12$

$ \text { Total }=12+12+6+18+12=60 $

$\therefore$ The number of numbers which are divisible by 4 are 60.

Given that the number of circular permutation of 9 distinct things taken 5 at a time is $n_1$.

$ \begin{aligned} \therefore \quad n_1 & ={ }^9 C_5(5-1)! \\\\ & =\frac{9!}{5!(9-5)!} \times 4!=9 \times 8 \times 7 \times 6 \end{aligned} $

Also, given that the number of linear permutation of 8 distinct things taken 4 at a time is $n_2$.

$ \begin{aligned} \therefore n_2={ }^8 C_4(4!) & =\frac{8!}{4!(8-4)!} \times 4! \\\\ & =8 \times 7 \times 6 \times 5 \end{aligned} $

Now, $\frac{n_1}{n_2}=\frac{9 \times 8 \times 7 \times 6}{8 \times 7 \times 6 \times 5}=\frac{9}{5}$

We know that

The number of ways in which $n$ different things can be distributed to $m$ person so that no person gets all the things is same as the number of functions (whose range is not a singleton set) from set $A$ (containing $n$ elements) to set $B$ (containing $m$ elements).

$\therefore$ The number of ways in which 4 different things can distributed to 6 person so that no person gets all the things

$ \begin{aligned} & =6^4-6=6(216-1) \\\\ & =6(215)=1290 \end{aligned} $

We have,

$ 2,3,5,7 $

No repetition

Number of given digits $=n=4$

Sum of digits = 17

Sum of all four digits formed

$ \begin{aligned} & =(n-1)!(\text { Sum of digits) (11111 ......... n times) } \\\\ & =3!(17)(1111) \\\\ & =6 \times 17 \times 1111 \\\\ & =113322 \end{aligned} $

Given that there are 15 identical gold coins to be distributed among 3 persons.

Also, each person gets at least 3 coins $\Rightarrow$ total $=3 \times 3=9$ coins

$\therefore 9$ coins are already distributed in 1 ways. we now have 6 coins to be distributed between 3 people.

$ \begin{aligned} & \Rightarrow x_1+x_2+x_3=6, x_i \geq 0 \\\\ & \text { Number of } \end{aligned} $

Number of solutions is given by

$ \begin{aligned} & { }^{n+r-1} C_{r-1} \\\\ & \Rightarrow \quad n=6, r=3 \\\\ & \therefore \text { Reanired solution } \end{aligned} $

$\therefore$ Required Number $={ }^{6+3-1} C_{3-1}$

$ ={ }^8 C_2=28 $

4 letter words from

ACCOMMODATION

There are different letters

"ACCOMMODATION"

namely,

A, C, D, I, M, N, O, T

$\begin{array}{llllllll}2 & 2 & 1 & 1 & 2 & 1 & 3 & 1\end{array}$

Case (I) all 4 letters of the word are different numbers of combinations

$ ={ }^8 c_4=70 $

Case (II) 2 letters are same and 2 are different.

$ ={ }^4 c_1 \cdot{ }^7 c_2-6=78 $

( $\because$ when we select 2 same letter from 0 and 3rd letter also from, 0 then this need to be subtracted and such cases are 6)

Case (III) When 3 letters are same

$ =1 .{ }^7 c_1=7 $

Case (IV) When 2 letters are same and other 2 are also same.

$ ={ }^4 c_1 \times{ }^3 c_1=12 $

Hence, total number of combination

$ =70+78+7+12=167 $

We have,

$ \begin{aligned} & 2 \frac{c_1}{c_0}+4 \frac{c_2}{c_1}+6 \frac{c_3}{c_2}+\ldots .+2 n \frac{c_n}{c_{n-1}}=650 \\\\ & \Rightarrow \quad \sum_{r=1}^n 2 r \frac{{ }^n c_r}{{ }^n c_{r-1}}=650 \end{aligned} $

$\begin{aligned} & \Rightarrow \quad 2 \sum_{r=1}^n r\left[\frac{\frac{n!}{r!(n-r)!}}{\frac{n!}{(r-1)!(n-r+1)!}}\right]=650 \\\\ & \Rightarrow 2 \sum_{r=1}^n r\left[\begin{array}{l}\frac{n!}{r \cdot(r-1)!(n-r)!} \\\\ \times \frac{(r-1)!(n-r+1) \cdot(n-r)!}{n!}\end{array}\right]\end{aligned}$

$ \begin{array}{ll} =650 \\\\ \Rightarrow & 2 \Sigma(n-r+1)=650 \\\\ \Rightarrow & 2\left[n^2+n-\frac{n(n+1)}{2}\right]=650 \\\\ \Rightarrow & n^2+n-650=0 \\\\ \Rightarrow & (n+26)(n-25)=0 \\\\ \Rightarrow & n=25 \end{array} $

Hence, ${ }^n c_2={ }^{25} c_2=300$

Case I When the unit digit is either 0 or 4 . Then, this digit should be 0,2 or 4.

$\therefore 4$-digit number in case 1 is

$ =4 \times 5 \times 3 \times 2=120 $

Case II When, the unit digit is 2 . Then, the ten's digit is either 1 or 3 .

$\therefore 4$-digit number in case $2=4 \times 5 \times 2 \times 1$ $=40$

Hence, required number

$ =120+40=160 . $

To determine the number of ways to arrange 2 red roses, 3 white roses, and 5 yellow roses of different sizes into a garland, ensuring that no two yellow roses are adjacent, we follow these steps:

Arrange Red and White Roses:

Since a garland has no fixed starting point, arranging the 2 red and 3 white roses in a circular fashion is equivalent to arranging them linearly with a rotational symmetry. This can be calculated using the formula for circular permutations, adjusted for identical elements:

$ \frac{(5-1)!}{2} = \frac{4!}{2} $

Calculate Available Spaces:

After arranging the 2 red and 3 white roses, we create "gaps" between them and at the ends. This creates 5 gaps.

Arrange Yellow Roses:

We now need to place the 5 yellow roses in these 5 gaps without any two yellow roses being adjacent, which can be done in $5!$ ways (as we are using all available spaces).

The calculation for this arrangement is done using permutations:

$ ^5P_5 = 5! $

Total Arrangements:

The total number of ways to arrange the garland is the product of ways to arrange the red and white roses and the distinct ways to fill the gaps with yellow roses:

$ \frac{4!}{2} \times ^5P_5 = \frac{4! \times 5!}{2} $

Calculation:

Calculate the factorials:

$ 4! = 24, \quad 5! = 120 $

Calculate the total number of arrangements:

$ \frac{24 \times 120}{2} = 1440 $

Thus, the required number of ways to arrange the roses into a garland such that no two yellow roses are adjacent is 1440.

Let there be $n$ men participents, then the number of games that the men play between themselves is $2 \times{ }^n C_2$ and the number of games that the men played with the women is $2 \times 2 n$

$\therefore 2 \times{ }^n C_2-2 \times 2 n=66\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,[given]$

or $n^2-5 n-66=0 \Rightarrow n=11$

Hence, number of participents men and women $=11$ men +2 women $=13$

The number of ways arranging a men around a circular table $=8!$

We have, 9 places to sit of 5 women $={ }^9 P_5$

$\therefore$ Number of ways $=8!\cdot{ }^9 P_5$

Since the fruits are alike, you can choose any number of fruits from 1 up to 6. This gives $6$ possible selections (choosing 1, 2, 3, 4, 5, or 6 fruits).

For the vegetables, you can choose any number from 1 up to 7, which gives $7$ possible selections.

For the biscuits, you can choose any number from 1 up to 8, providing $8$ possible selections.

Since these choices are made independently (one choice does not affect the others), you multiply the number of ways in each category:

$\text{Total number of ways} = 6 \times 7 \times 8.$

Calculating the product:

$6 \times 7 = 42,$

$42 \times 8 = 336.$

Thus, the number of ways of selecting at least one fruit, one vegetable, and one biscuit is $\boxed{336}$.

The correct answer, therefore, is Option B.

Write the word TABLE

alphabetically A, B, E, L, T

To finding the rank of BLATE

Fixing B, arrangements with $A$ in the

first position $4!=24(A+$ permutations

of $B, E, L, T$ )

If permutation starting with $B$, then

fixing $B$ and then $L$, arrangements of,

$A, B, E, T: 3!=6(B L+$ permutations of

$A, B, E, T$ )

if permuatations starting with $B L$

Fixing BL, arrangements of A, B, E, T

$ =3!=6 $

(BLA + permutations)

arrangements of $A, E, B, T: 2!=2$

(BLAE + permutations)

arrangement of $A, E, T: 1!=1$

(BLATE + permutations)

Thus, fixing $A=24$

fixing $B=A+6=30$

Fixing L: BLA $=6$

Final E: BLAE $+1=31$

Therefore, total permutations of BLATE excluding irredevent permutations.

Thus, BLATE +31 : The possible rank of Table are permutations $=30+31=61$

Linear arrangements where no two boys sit together (X):

First, arrange the 6 girls. This can be done in $5!$ (120) ways since the arrangement of the boys around these gaps is what's crucial. Once the girls are arranged, they create 7 possible gaps (1 before each girl and 1 after the last girl) to place the boys.

Next, select 5 of these 7 gaps to place a boy. This can be done in ${}^7C_5$ ways. Each selection of gaps can be filled by arranging the boys in $5!$ different ways.

Therefore, the total number of linear arrangements where no two boys sit together (X) is:

$ X = 5! \times {}^7C_5 \times 5! = 5! \times \frac{7!}{5!2!} \times 5! $

Simplifying,

$ X = \frac{7 \times 6}{2} \times 5! \times 5! = 21 \times 5! \times 5! $

Linear arrangements where no two girls sit together (Y):

This problem can be solved similarly to the arrangement for X. Here, arrange the boys first, which can also be done in $5!$ ways. This arrangement creates 6 gaps.

All 6 girls can be placed into these exact 6 gaps in ${}^6C_6$ ways, and then arrange those girls among themselves in $6!$ (720) ways.

Thus, the total number of linear arrangements where no two girls sit together (Y) is:

$ Y = 5! \times {}^6C_6 \times 6! = 5! \times 1 \times 6! $

Since $Y = 5! \times 6!$, and recognizing $6! = 6$, it remains:

$ Y = 5! \times 5! $

Circular arrangements where no two boys sit together (Z):

For arranging 11 people in a circle, consider fixing one girl as a reference point, leaving 10 people to arrange, which alters our calculation by one less arrangement over linear. Arrange the remaining 5 boys and 5 girls around this fixed position.

The number of circular arrangements, where no two boys sit together for them is:

$ Z = 5! \times 5! $

For the final result, calculating the ratio $X:Y:Z$ gives us:

$ X:Y:Z = 21 \times 5! \times 5! : 1 \times 5! \times 5! : 1 \times 5! \times 5! = 21 : 1 : 1 $

Distribute the Minimum Apples:

$ A $ receives 2 apples, and $ C $ receives 2 apples.

This leaves $ 15 - 2 - 2 = 11 $ apples to be distributed among $ A $, $ B $, and $ C $.

Define Variables for Remaining Distribution:

Let $ x $ be the number of apples $ A $ gets over the initial 2.

Let $ y $ be the number of apples $ B $ receives.

Let $ z $ be the number of apples $ C $ gets over the initial 2.

The equation to solve is $ x + y + z = 11 $.

Apply Constraints:

Since $ B $ can receive at most 5 apples, $ y \leq 5 $.

Use the Stars and Bars Method:

Calculate the total number of solutions to $ x + y + z = 11 $ without any constraints:

$ \binom{11 + 2}{2} = \binom{13}{2} = 78 $

Account for the Constraint on $ B $:

For cases where $ y > 5 $, set $ y' = y - 6 $, where $ y' \geq 0 $.

Substitute $ y $ with $ y' $ to get the equation $ x + y' + z = 5 $.

Count the number of solutions to $ x + y' + z = 5 $:

$ \binom{5 + 2}{2} = \binom{7}{2} = 21 $

Calculate the Final Result:

Subtract the invalid cases from the total solutions:

$ 78 - 21 = 57 $

Thus, there are 57 ways to distribute the apples based on the given conditions.

and number of poetry $=3$

$\therefore$ Number of ways of selecting 4 novels from 6

$ ={ }^6 C_4 \text { way } $

Number of ways of selecting 1 poetry from 3.

$ ={ }^3 C_1 \text { way } $

When poetry is always in middle,

So, 4 nowels can be arrange in 4 ! ways.

$\therefore$ Required number of ways

$ \begin{aligned} & ={ }^6 C_4 \times{ }^3 C_1 \times 4! \\ & =\frac{6 \times 5}{2 \times 1} \times 3 \times 24 \\ & =15 \times 3 \times 24 \\ & =1080 \end{aligned} $

To form a five-digit number using the digits 0, 1, 2, 3, 4, and 5, which must be divisible by 3, we need to consider the divisibility rule for 3. A number is divisible by 3 if the sum of its digits is a multiple of 3. This gives us two possible scenarios to examine:

Case I: Using the digits 0, 1, 2, 4, and 5.

The sum of these digits is $0 + 1 + 2 + 4 + 5 = 12$, which is divisible by 3.

To avoid starting with 0 (as it wouldn't be a five-digit number), the first digit has 4 options (1, 2, 4, or 5).

The remaining four digits can be arranged in $4!$ (4 factorial) ways.

Thus, the number of ways to arrange these digits is:

$ 4 \times 4 \times 3 \times 2 \times 1 = 96 $

Case II: Using the digits 1, 2, 3, 4, and 5.

The sum of these digits is $1 + 2 + 3 + 4 + 5 = 15$, which is also divisible by 3.

Any of these five digits can be at the start, providing 5 choices for the first digit.

The remaining four digits are arranged in $4!$ ways.

Thus, the number of ways to arrange these digits is:

$ 5 \times 4 \times 3 \times 2 \times 1 = 120 $

Conclusion:

Adding the numbers from both scenarios gives the total number of five-digit numbers divisible by 3:

$ 96 + 120 = 216 $

Four-digit number formed using the digits $\{1,2,3,4,5,6,7\}$ with all distinct digits.

$ p={ }^7 P_4=7 \times 6 \times 5 \times 4=840 $

Case I Leading digit is 3 .

For the number to be greater than 3400 , the second digit must be $4,5,6$ or 7 .

(i) Fix 3 as first digit.

(ii) Choose one of 4, 5, 6 or 7 for the second digits ( 4 choices).

(iii) For the third and fourth digits, Choose from the remaining 5 digits So, for each choice of the second digit $5 \times 4=20$

There are 4 choices for the second digits so, $4 \times 20=80$

Case II Leading digit is $4,5,6$ or 7 .

Any number starting with $4,5,6$ or 7 is automatically greater than 3400 .

There are 4 choices for the first digit (4,5,6 or 7 )

Choose any 3 digits from the remaining 6 digits.

So, for each choice of the first digit

$ 6 \times 5 \times 4=120 $

There are 4 choices for the first digit, so

$ 4 \times 120=480 $

Total numbers greater than 3400

$ q=80+480=560 $

Now, $\frac{p}{q}=\frac{840}{560}=\frac{3}{2}$

So, the ratio $p: q=3: 2$.

In all the numbers unit digit either 3 or 5 or 7 .

All 5 digit number (with repetition)

$ \begin{aligned} & \{4,5,6,7\}\{0,3,4,5,6,7\}\{0,3,4,5,6,7\} \\ & \{0,3,4,5,6,7\}\{3,5,7\} \\ & =4 \times 6 \times 6 \times 6 \times 3=2592 \end{aligned} $

All 5 digit number (without repetition)

$ \begin{aligned} & \text { Case } I \frac{}{\{4,5,6,7\}}---\frac{3}{}= 4 \times 4 \times 3 \times 2 \times 1=96 \\\\ & \text { Case II } \frac{}{\{4,6,7\}}---\frac{5}{}=3 \times 4 \times 3 \times2 \times 1=72\\ & \text { Case III } \frac{}{\{4,5,6\}}---\frac{7}{}=3 \times 4 \times 3 \times 2 \times 1=72\\ & \text { Total numbers (without repetition) } \\ & =96+72+72=240 \\ & \therefore \text { Required number }=2592-240=2352 \end{aligned} $

To find the number of ways to arrange 3 men and 3 women in a row of 6 seats such that the first and last seats are occupied by men, follow these steps:

Arrangement Patterns

Since the first and last seats must be occupied by men, we have two initial positions occupied. After placing two men in these spots, we are left with 1 man and 3 women to arrange in the remaining 4 seats. Here are the possible patterns:

Men, Women, Men, Women, Women, Men

Men, Women, Women, Men, Women, Men

Men, Men, Women, Women, Women, Men

Men, Women, Women, Women, Men, Men

Calculating the Number of Arrangements

For each pattern:

Men: We have 3 men, and we've used 2 positions (first and last), leaving us with 1 man to arrange. This can be done in $3!$ ways.

Women: With 3 women, we fill the remaining positions, which can also be done in $3!$ ways.

Since there are 4 patterns, the total number of arrangements is calculated as:

$ = 4 \times (3!) \times (3!) $

Breaking down the calculation:

$3! = 3 \times 2 \times 1 = 6$

So, using these numbers, the calculation is:

$ = 4 \times 6 \times 6 = 144 $

Thus, there are 144 different ways to arrange 3 men and 3 women in a row of 6 seats, with the first and last seats filled by men.

For majority, Men should be $>$ ifl number.

Selecting the committee

$ \Rightarrow{ }^8 C_6 \times{ }^6 C_4+{ }^8 C_7 \times{ }^6 C_3+{ }^8 C_3 \times{ }^6 C_2 $

$ \begin{aligned} &\Rightarrow \quad \frac{8 \times 7 \times 6 \times 5}{2 \times 1 \times 2 \times 1}+\frac{8 \times 6 \times 5 \times 4}{1 \times 3 \times 2 \times 1}+\frac{8 \times 6 \times 5}{8 \times 2 \times 1} \end{aligned} $

$ =420+160+15=595 $

As there are 4 choices for each question, so we can answer the question in $(4 \times 4 \times 4)$ ways.

And there lies one way to answer the question in $(4 \times 4 \times 4)$, ways to get all the correct answer which should be neglected.

$\therefore$ Maximum possible number of students $=(4 \times 4 \times 4-1)=\left(4^3-1\right)=63$

A number between 1000 and 10000 contains 4 -digits So, we have to form 4-digit numbers having exactly two of their digits as 3 and 7 only once.

So, we have the following ways to form

4 - digit numbers with above conditions

I. $7 \times 8 \times 1 \times 1=56$ numbers

II. $7 \times 1 \times 1 \times 8=56$ numbers

III. $1 \times 1 \times 8 \times 8=64$ numbers

IV. $1 \times 8 \times 1 \times 8=64$ numbers

V. $1 \times 8 \times 8 \times 1=64$-numbers

VI. $7 \times 8 \times 1 \times 1=56$ numbers

Thus, total number of required type of numbers

$ =2 \times(56+56+64+64+64+56) \text { numbed } $

( $\because 3$ and 7 can be interchanged.)

$=2 \times 360$ numbers

$=720$ numbers

The general formula for distribuind $n$ things amongest $r$ people is

$ { }^{n-1} C_{r-1}={ }^{17-1} C_{4-1}={ }^{16} C_3=560 $

and the number of ways to distributel objects into $k$ groups such that each group at least $m$ object is ${ }^{n-k+m} C_{k-1}$

where $n=17, k=$ number of groups $=4$ $m=$ minimum number of objects $=3$ $\begin{aligned}{ }^{17-4 \times 3} C_{4-1} & ={ }^{17-12} C_3 \\ & ={ }^5 C_3=10\end{aligned}$

$ ={ }^5 C_3=10 $

$\therefore$ Required number of ways

$ =\frac{560}{10}=56 $

To find the number of sides $ n $ of a polygon that has 275 diagonals, we use the formula for the number of diagonals in an $ n $-sided polygon:

$ \text{Number of diagonals} = \frac{n(n-3)}{2} $

Given that there are 275 diagonals, we set up the equation:

$ \frac{n(n-3)}{2} = 275 $

Multiplying both sides by 2 to eliminate the fraction gives:

$ n(n-3) = 550 $

This expands to the quadratic equation:

$ n^2 - 3n - 550 = 0 $

To solve for $ n $, we use the quadratic formula:

$ n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $

where $ a = 1 $, $ b = -3 $, and $ c = -550 $. Plugging these values into the formula:

$ n = \frac{-(-3) \pm \sqrt{(-3)^2 - 4 \times 1 \times (-550)}}{2 \times 1} $

Simplifying within the square root:

$ n = \frac{3 \pm \sqrt{9 + 2200}}{2} $

$ n = \frac{3 \pm \sqrt{2209}}{2} $

The square root of 2209 is 47, so:

$ n = \frac{3 \pm 47}{2} $

This gives us two potential solutions:

$ n = \frac{3 + 47}{2} = \frac{50}{2} = 25 $

$ n = \frac{3 - 47}{2} = \frac{-44}{2} = -22 $

Since $ n $ must be a positive integer, the number of sides in the polygon is 25.

$ \begin{array}{l|l} 2 & 1080 \\ \hline 2 & 540 \\ \hline 2 & 270 \\ \hline 3 & 135 \\ \hline 3 & 45 \\ \hline 3 & 15 \\ \hline 5 & 5 \\ \hline & 1 \end{array}$

$ 1080=2^3 \times 3^3 \times 5^4 $

Total number of positive divisors

$ =\left(e_1+1\right)\left(e_2+1\right) \ldots\left(e_n+1\right) $

Where $e_1, e_2, \ldots . e_n$ are the exponents in the prime factorisation.

so, number of positive divisors of

$ \begin{aligned} 1080 & =(3+1)(3+1)(1+1) \\ & =4 \times 4 \times 2=32 \end{aligned} $

We have, $a_n=\sum\limits_{r=0}^n \frac{1}{n_{c_r}}$

Let $S=\sum\limits_{r=0}^n \frac{r}{{ }^n c_r}$

Replacing $r$ by $n-r$, we get

$ \begin{aligned} & S=\sum\limits_{r=0}^n \frac{n-r}{{ }^n c_{n-r}} \\ & S=\sum\limits_{r=0}^n \frac{n-r}{{ }^n c_r} \end{aligned} $

$ \text { [as }{ }^n c_r={ }^2 c_{n-1} $

On adding Eqs. (i) and (ii), we get

$ \begin{aligned} & 2 S=\sum_{r=0}^n \frac{r}{n_{c_r}}+\sum_{r=0}^n \frac{n-r}{{ }^n c_r} \\ & 2 S=n \sum_{r=0}^n \frac{1}{n^{c_r}} \\ & 2 S=n \cdot a_n \Rightarrow S=\frac{n}{2} \cdot a_n \end{aligned} $

The alphabetical order of the letters of the given word MASTER is A, E, M, R, S and T.

Number of ways begin with letter $A$

$ =5!=120 $

Number of ways begin with letter E

$ =5!=120 $

Number of ways begin with letter MAE

$ =3!=6 $

Number of ways being with letter MAR

$ =3!=6 $

Number of ways begin with letter

$ \text { MASE }=2!=2 $

Number of ways begin with letter

$ \text { MASR }=2!=2 $

Next word is MASTER.

$\therefore$ Rank of the word MASTER in the dictionary order

$ =120+120+6+6+2+2+1=257 $

Total number of elements $=8$

$\therefore$ Required number of subsets

$\begin{aligned} & ={ }^8 C_6+{ }^8 C_7+{ }^8 C_8=\frac{8 \times 7}{1 \times 2}+8+1 \\ & =28+9=37\end{aligned}$

There are total 10 letters in the word 'REPETITION'.

Case I There are 7 distinct letter in the given work all 4 letters are distinct.

So, number of ways of choosing 4 letters out of $7={ }^7 C_4$.

4 letters can be arranged in 4 ! ways.

$\therefore$ Number of ways of arranging these 4 letters $={ }^7 C_4 \times 4!=840$ ways

Case II Two distinct and 2 alike letters number of ways of choosing two distinct and 2 alike letters.

$ =\frac{{ }^3 C_1 \times{ }^6 C_2 \times 4!}{2!}=3 \times 15 \times 12=540 $

Case III Both pair are alike.

Number of ways of choosing both alike pairs

$ =\frac{{ }^3 C_2 \times 4!}{2!\times 2!}=18 $

$\therefore$ Required number of ways

$ =840+540+18=1398 $

To prepare a garland using 6 distinct white roses and 6 distinct red roses such that no two red roses are adjacent, follow these steps:

Arranging White Roses:

Since we have 6 distinct white roses, they can be arranged in a garland. The formula to arrange flowers in a garland using circular permutations is $\frac{(n-1)!}{2}$, where $n$ is the number of items. For the white roses, this gives us:

$ \frac{(6-1)!}{2} = \frac{5!}{2} = \frac{120}{2} = 60 $

Arranging Red Roses:

When the white roses are arranged, they create gaps in between them where red roses can be placed so that no two red roses come together. For 6 white roses arranged circularly, this results in 6 distinct positions for placing the red roses to separate them.

Placing Red Roses:

The number of ways to arrange these 6 distinct red roses in these 6 positions is given by the permutation of 6 distinct items, which is $6!$. Thus, the calculation is:

$ 6! = 720 $

Calculating Total Arrangements:

Finally, combine the permutations of both sets of roses. The total number of ways to arrange the garland while maintaining the condition is:

$ 60 \times 720 = 43200 $

Thus, there are 43,200 different ways to prepare the garland with the given conditions.

We have 10 men and 8 women.

Possible number of ways of forming a committee are

5 women, 3 men or 6 women, 2 men or 7 women, 1 men or 8 women

$\therefore$ Required number of ways

$ \begin{aligned} & ={ }^8 C_5 \times{ }^{10} C_3+{ }^8 C_4 \times{ }^{10} C_2+{ }^8 C_7 \times{ }^{10} C_1+{ }^4 C_1 \\ & =6720+1260+80+1=8061 \end{aligned} $

To find the rank of the word "CRICKET" when all its permutations are arranged in dictionary order, follow these steps:

List the Letters in Alphabetical Order: Arrange the letters of "CRICKET" alphabetically: C, C, E, I, K, R, T.

Compute Permutations for Each Case:

Starting with C followed by C (CC): Calculate permutations of the remaining letters E, I, K, R, T. This would be $ 5! = 120 $ ways.

Starting with C followed by E (CE): Calculate permutations of the remaining letters C, I, K, R, T. This would also be $ 5! = 120 $ ways.

Starting with C followed by I (CI): Calculate permutations of the remaining letters C, E, K, R, T. Again, $ 5! = 120 $ ways.

Starting with C followed by K (CK): Calculate permutations of the remaining letters C, E, I, R, T. Again, $ 5! = 120 $ ways.

Starting with C followed by R and then C (CRC): Calculate permutations of the remaining letters E, I, K, T. This is $ 4! = 24 $ ways.

Starting with C followed by R and then E (CRE): Calculate permutations of the remaining letters C, I, K, T. This also results in $ 4! = 24 $ ways.

Starting with CRIC: You need to arrange the remaining letters K, E, T. "CRICE" results in 2 distinct permutations:

CRICEK

CRICKET (the word we are interested in)

Therefore, the number of arrangements before CRICKET is 2 (starting at CRICE_).

Determine the Rank:

The rank of "CRICKET" is therefore $ 120 + 120 + 120 + 120 + 24 + 24 + 2 + 1 = 531 $.

Thus, the rank of the word "CRICKET" is 531 in the list of all possible permutations arranged in dictionary order.