Permutations and Combinations

Calculate the number excluding the digit 5 .

(a) 1-digit number

$ =8(1,2,3,4,6,7,8,9) $

Similarly,

(c) 3-digit numbers $=8 \times 9 \times 9=648$

(d) 4-digit numbers $=8 \times 9 \times 9 \times 9=5832$

Total numbers between 1 and 10000 excluding the digit 5

$ =8+72+648+5832=6560 $

Hence, required numbers $=9999-6560$

$ =3439 $

Total number of people $=5+4=9$

So, total arrangements $=9$ !

To find $\alpha$ Let the particular man and woman be treated as a single unit or block when they are together.

Remaining 7 peoples +1 block $=8$ units

$ \therefore \quad \alpha=2 \times 8! $

and $\beta=9$ ! $-\alpha=9$ ! $-2 \times 8$ !

Hence, $\alpha: \beta=\frac{2 \times 8!}{8}: \frac{9!-2 \times 8!}{8!}$

$ =2: 7 $

Number of ways to choose 4 couples out of $4={ }^4 C_4=1$

From 4 couples, pick 1 person per couple $=2^4=16$

Team formula married couples $=2^2=4$

Hence, required ways $=16-4=12$

Total 8-digit numbers formed by $1,2,3,4,5,6,7,8$ is 8 !

Total 8 -digit numbers formed by $0,2,3$, $4,5,6,7,9$ is $7 \times 7$ !

Total 8 -digit numbers formed by $1,0,3$, $4,5,6,8,9$ is $7 \times 7$ !

Total 8 -digit numbers formed by $1,2,0$, $4,5,7,8,9$ is $7 \times 7$ !

Total 8-digit numbers formed by $0,1,2$, $3,5,6,7,8$ is $7 \times 7$ !

$ \begin{aligned} \text { Total } & =8!+28 \times 7! \\ & =(8+28) 7! \\ & =36 \times 7! \end{aligned} $

Given word, MATHEMATICS

$ \begin{array}{ll} 2 \mathrm{M} & 1 \mathrm{H} \\ 2 \mathrm{~A} & 1 \mathrm{E} \\ 2 \mathrm{~T} & 1 \mathrm{I} \\ & 1 \mathrm{C} \\ & 1 \mathrm{~S} \end{array} $

Number of words formed using 4 letters in which two are of same kind and 2 are different

$ \begin{aligned} & ={ }^3 C_1 \times{ }^7 C_2 \times \frac{4!}{2!} \\ & =3 \times 7 \times 3 \times 4 \times 3 \\ & =27 \times 28 \\ & =756 \end{aligned} $

Digits $-0,5,6,7,8$ and 9

No. greater than 6000

Case I 4 digit numbers greater than 6000

$ \begin{aligned} & \bar{\uparrow} \bar{\uparrow} \bar{\uparrow} \bar{\uparrow}=240 \\ & 4543 \end{aligned} $

$ \begin{aligned} &\text { Case II 5-digit numbers }\\ &\begin{aligned} & \bar{\uparrow} \bar{\uparrow} \bar{\uparrow} \bar{\uparrow} \bar{\uparrow}=600 \\ & 55432 \end{aligned} \end{aligned} $

Case III 6-digit numbers

$ \begin{aligned} & \bar{\uparrow} \bar{\uparrow} \bar{\uparrow} \bar{\uparrow} \bar{\uparrow} \bar{\uparrow}=600 \\ & 554321 \end{aligned} $

∴ Total $=1440$

∴ Required case

$=(2$ particular are in Ist group $)+$ (2 particular are in 3rd group) + (one is in Ist group and other is in 3rd group)

$ \begin{aligned} & =\frac{13!}{1!5!7!}+\frac{13!}{3!5!5!}+\frac{13!}{2!5!6!}(24) \\ & =\frac{13!}{7!5!}+\frac{13!}{6!5!}+\frac{13!}{5!6!}=\frac{13!(1+7+7)}{7!5!} \\ & =\frac{13 \times 12 \times 11!\times 15}{7!\times 5 \times 4 \times 3!}=\frac{117(11!)}{7!3!} \end{aligned} $

The digit are 1 to 9 . Each such number is strictly increasing sequence of digits, i.e., digits don't repeat and are in order.

e.g., 12, 135, 1239 and soon.

So, total such numbers in the number of increasing digit sequence of length 2 to 4.

We choose subsets of digits from 1 to 9 of lengths 2,3 and 4 in increasing order.

$\therefore 2$-digits numbers

$ ={ }^9 C_2=\frac{9!}{2!7!}=\frac{9 \times 8}{2}=36 $

3-digits numbers

$ ={ }^9 C_3=\frac{9!}{3!6!}=\frac{9 \times 8 \times 7 \times 6!}{6 \times 6!}=84 $

$ \begin{aligned} &\begin{aligned} 4 \text {-digits numbers } & ={ }^9 C_4=\frac{9!}{4!5!} \\ & =\frac{9 \times 8 \times 7 \times 6 \times 5!}{4 \times 3 \times 2 \times 5!}=126 \end{aligned}\\ &\text { So, total such digits }\\ &\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=36+84+126=246 \end{aligned} $

Given, word is AGAIN,

So the letters in alphabatical order : A, A, G, I, N

Words starting with $A=\frac{4!}{2!}=12$

Words starting with $G=\frac{4!}{2!}=12$

Words starting with $I=\frac{4!}{2!}=12$

Words starting with $N=\frac{4!}{2!}=12$

So, total words $=12+12+12+12=48$

Thus, the 50th word starts with $N$.

49th word = NAAGI

$\therefore 50$ th word $=$ NAAIG.

We select exactly I wicket keeper from 4 .

So, number of ways $={ }^4 C_1=4$

Now, we have 10 players left to select from batsmen, bowlers and all-rounders with the condition.

Batsmen $\geq 4$, Bowlers $\geq 3$, All-rounders $\geq 2$

So, total of these three $=10$

So, number of batsmen, $B$ from 4 to 6 , number of bowlers, $L$ (say)from 3 to 6 and number of all-rounder, $A$ (say) from 2 to 4 .

$\therefore B+L+A=10$ (after 1 wicket keeper)

Now, the possible combinations for, (B, L, A) are

$(4,4,2),(4,3,3),(5,3,2)$

Total number of ways for 10 players

$ \begin{aligned} & =\left({ }^6 C_4{ }^6 C_4{ }^4 C_2+{ }^6 C_4{ }^6 C_3{ }^4 C_3+{ }^6 C_5{ }^6 C_3{ }^4 C_2\right) \\ & =(15 \times 15 \times 6+15 \times 20 \times 4+6 \times 20 \times 6) \\ & =(1350+1200+720)=3270 \end{aligned} $

$\therefore$ Total number of ways for 11 players

$ \begin{aligned} & =4 \times 3270 \\ & =13080 \end{aligned} $

Number of combinations of choosing 4 digits out of 4

$ ={ }^5 C_4=5 $

Each 4-digit set can be arranged in 4! ways $=24$ ways

So, total number of 4 -digit numbers

$ =5 \times 24=120 $

In each group, each digit appears

$ \frac{24}{4}=6 \text { times } $

So, contribution per group

$ =\text { sum of digits } \times 6 \times(1000+100+10+1) $

Now, compute total sum of digits for all 5 groups

$ \begin{array}{r} \begin{array}{r} (2+3+5+8)+(1+3+5+8) \\ +(1+2+5+8)+(1+2+3+8) \\ +(1+2+3+5) \end{array} \\ =18+17+16+14+11=76 \end{array} $

$ \text { Hence, total sum } \begin{aligned} & =76 \times 6666 \\ & =506616 \end{aligned} $

Total possible ordered pairs $(p, q)$ of first 50 natural numbers

$ =50 \times 50=2500 $

Out of these, exactly half will satisfy $p>q$ and there are 50 pairs where $p=q \{(1,1),(2,2) \ldots(50,50)\}$

So, number of pairs with $p \neq q$

$ =2500-50=2450 $

and for $p>q$

$ =\frac{2450}{2}=1225 $

Fathers $(F)=5$, Teachers $(T)=6$, students $(S)=4$

for, $T+F+S=7, T \geq 0$, thus $F \geq 1, S \geq 1$ total combination ( $T, F, S$ )

$ (4,1,2),(4,2,1),(5,1,1),(3,2,2) $

Case $\mathbf{I}(T=4, F=1, S=2)$

$ \begin{aligned} & ={ }^6 C_4 \cdot{ }^5 C_1 \cdot{ }^4 C_2 \\ & =15 \times 5 \times 6=450 \end{aligned} $

$ \text { } \begin{aligned} \mathbf{Case \,I I}\,\,(T & =4, F=2, S=1) \\ & ={ }^6 C_4 \cdot{ }^5 C_2 \cdot{ }^4 C_1 \\ & =15 \times 10 \times 4=600 \end{aligned} $

Case III ( $T=5, F=1, S=1$ )

$ \begin{aligned} & ={ }^6 C_5 \cdot{ }^5 C_1 \cdot{ }^4 C_1 \\ & =6 \times 5 \times 4=120 \end{aligned} $

Case IV ( $T=3, F=2, S=2$ )

$ \begin{aligned} & ={ }^6 C_3 \cdot{ }^5 C_2 \cdot{ }^4 C_2 \\ & =20 \times 10 \times 6=1200 \end{aligned} $

Hence, total number of ways

$ \begin{aligned} & =450+600+120+1200 \\ & =2370 \end{aligned} $

The total number of ways that all sisters and brothers seated around a circle is 10 !

The number of ways that the three sisters seated together along with the brothers is $8!\times 3$ !

So, required number of ways

$ =10!-8!3!=8!(10 \times 9-6)=8!\times 84 $

Given, 8 students in a class room 4 of them are chosen and arrange in ground table and 4 in row

$ \begin{aligned} & \therefore \text { Total arrangement }={ }^8 C_4(3!+4!) \\ & \quad=\frac{8!}{4!4!}(6+24)=70 \times 30 \\ & \quad=2100 \end{aligned} $

Give word VARIABLE

A twice and other are different arrangement of three letter words.

Case 12 alike 1 distinct

$ ={ }^1 C_1 \times{ }^6 C_1 \times \frac{3!}{2!}=6 \times 3=18 $

Case II All are distinct

$ \begin{aligned} { }^7 C_3 \times 3! & =35 \times 6=210 \\ \therefore \text { Total } & =210+18=288 \\ n(S) & =288 \end{aligned} $

Middle letter is consonant.

$ =6 \times 4 \times 5=120 $

$ =1 \times 4=4 $

Total favourable case $=120+4=124$

$\therefore$ Required probability $=\frac{124}{288}=\frac{31}{57}$

$ \begin{aligned} &\\ &\begin{aligned} & \text { }{ }^{27} P_{r+7}=7722{ }^{25} P_{r+4} \\ & \Rightarrow \frac{27!}{20-r!}=7722 \cdot \frac{25!}{21-r!} \\ & \Rightarrow \frac{27 \cdot 26 \cdot 25!}{20-r!}=7722 \cdot \frac{25!}{(21-r) 20-r!} \\ & \Rightarrow 27 \cdot 26=7722 \times \frac{1}{(21-r)} \\ & \Rightarrow \text { By option put } r=10 \\ & \Rightarrow \quad 702=702 \end{aligned} \end{aligned} $

Let number of side $=n$

The number of diagonals of a regular

$ \begin{array}{ll} \Rightarrow & \frac{n(n-3)}{2}=35 \\ \Rightarrow & n^2-3 n=70 \\ \Rightarrow & n^2-3 n-70=0 \\ \Rightarrow & n^2-10 n+7 n-70=0 \\ \Rightarrow & n(n-10)+7(n-10)=0 \\ \Rightarrow & (n-10)(n+7)=0 \\ \Rightarrow & n=10, n=-7(\text { neglected }) \\ \therefore & n=10 \end{array} $

ASSIGNMENT

Total $=10$

2 S and 2 N

Different $=8$

A, I, G, M, E, T

Case I All are different

$ ={ }^8 C_4 \times 4!=1680 $

Case II Two are same and two are different

$ \begin{aligned} & ={ }^2 C_1 \times{ }^7 C_2 \times \frac{4!}{2!}=756 \\ & =2!\times \frac{7!}{2!\cdot 5!} \times \frac{4!}{2!} \\ & =\frac{7 \times 6 \times 5!}{2!\times 5!} \times 4 \times 3 \times 2!=504 \end{aligned} $

Case III Two are same and two are same

$ ={ }^2 C_2 \times \frac{4!}{2!\times 2!}=6 $

Total $=1680+504+6=2190$

We have, $\mathrm{E}, \mathrm{E}, \mathrm{L}, \mathrm{R}, \mathrm{T}, \mathrm{T}$

Words starting with $\mathrm{E}, \mathrm{L}, \mathrm{R}$, where first letter is T

(i) Starting with $\mathrm{E}: \frac{5!}{2!}=60$

(ii) Starting with $\mathrm{L}: \frac{5!}{2!2!}=30$

(iii) Starting with $\mathrm{R}: \frac{5!}{2!2!}=30$

Total $=60+30+30=120$

Fixed TE next is T

Letters before T is E, L, R

Each gives $3!=6$

$ \Rightarrow 3 \times 6=18 $

Now, fixed TETL, next is E one word before TETLE

$k$ is remaining letter which is before 2 letters.

Final letter is TETLER =1

Rank $=120+18+2+1=141$.

Given digits $0,1,2,3,5,7$

To find rank of 70513

Total number start with $1=5!=120$

Number start with $2=5!=120$

Number start with $3=5!=120$

Number start with $5=5!=120$

Number start with $701=6$

Number start with $702=6$

Number start with $703=6$

$ \begin{aligned} & 70512=1 \\ & 70513=1 \end{aligned} $

Rank of 70513

$ \begin{aligned} & =4 \times 120+3 \times 6+2 \\ & =500 \end{aligned} $

Number of divisor of 7!.

$ \begin{aligned} 7! & =7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 \\ & =2^4 \times 3^2 \times 5^1 \times 7^1 \end{aligned} $

Number of divisor

$ \begin{aligned} & =(4+1)(2+1)(1+1)(1+1) \\ & =60 \end{aligned} $

COMBINATION

Vowels- O, O, I, I, A = 5

$ \begin{aligned} &=4 \times \frac{8!}{2!2!} \times 2(\mathrm{C} \text { and } \mathrm{N} \text { are also be arranged })\\ &=2 \times 8! \end{aligned} $

Total number of fruits $=3 \times 12=36$

These fruits distributed among 9 persons equals $\because$ Order is taken'.

Total number of ways $=\frac{36!}{(4!)^9 9!} \times 9!=\frac{36!}{(4!)^9}$.

We have, $\sum\limits_{i=0}^m{ }^{10} C_i{ }^{20} C_{m-i}$

$ ={ }^{10} C_0{ }^{20} C_m+{ }^{10} C_1{ }^{20} C_{m-1}+{ }^{10} C_2{ }^{20} C_{m-2}+\ldots+{ }^{10} C_m \cdot{ }^{20} C_0 $

Coefficient of $x^m$ in the expansion of the product

$ (1+x)^{10}(1+x)^{20} $

$=$ Coefficient of $x^m$ in $(1+x)^{30}$

$ ={ }^{30} C_m . $

Maximum value of ${ }^{30} C_m={ }^{30} C_{15}$

$ \therefore \quad m=15 $

We have, $x y z=30$

$30=1,2,3,5,6,10,15,30$ are the integral factor of 30

$ \begin{aligned} & 30=3 \times 2 \times 5=3! \\ & 30=6 \times 5 \times 1=3! \\ & 30=10 \times 3 \times 1=3! \\ & 30=15 \times 2 \times 1=3! \\ & 30=30 \times 1 \times 1=\frac{3!}{2!} \end{aligned} $

$ \begin{aligned} \therefore \text { Total number of factor } & =4 \times 3!+\frac{3!}{2!} \\ & =24+3=27 \end{aligned} $

Given word is INEONVENIENCE. The distinet letters are I, N, C, O, V, E (6 letters) and their frequencies are

$ \begin{aligned} & \mathrm{I}-2 \\ & \mathrm{~N}-4 \\ & \mathrm{C}-2 \\ & \mathrm{O}-1 \\ & \mathrm{~V}-1 \\ & \mathrm{E}-3 \end{aligned} $

Total number of 5-letters words

(i) When 5 distinct letters, number of words $=6 P_5=720$

(ii) One pair and 3 distinct letters,

$ \begin{aligned} \text { number of words } & ={ }^4 C_1 \times{ }^5 C_3 \times \frac{5!}{2!} \\ & =4 \times 10 \times 60=2400 \end{aligned} $

(iii) One triple and 2 distinct letters, number of words

$ ={ }^2 C_1 \times{ }^5 C_2 \times \frac{5!}{3!} $

$ =2 \times 10 \times 20=400 $

(iv) Two pairs and one distinct, number

$ \begin{aligned} \text { of words } & ={ }^4 C_2 \times{ }^4 C_1 \times \frac{5!}{2!2!} \\ & =6 \times 4 \times 30=720 \end{aligned} $

(v) One triplet and one pair, number of words

$ =4 \times \frac{5!}{3!2!}=4 \times 10=40 $

(vi) One 4 a like and one different

$ ={ }^1 C_1 \times{ }^5 C_1 \times \frac{5!}{4!}=25 $

$\therefore$ Number of words with at least one repealed letters

$ \begin{aligned} & =(720+2400+400+720+40+25)-720 \\ & =3585 \end{aligned} $

Given, PERFECTION

Vowels E E I O and consonants PRFC T M.

Vowels can be arranged by $\frac{4!}{2!}$

$ \begin{aligned} & =4 \cdot 3!=2!3! \\ & E \times X E \times X \times X O \end{aligned} $

Consonants are arranged at $\times$ places by 6 ! ways.

$\because$ Total number of ways $=6!\times 3!\times 2!$

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

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