Permutations and Combinations

Given, $S=\{1,2,3,4,5,6,7,8,9\}$

1. Calculation for x

For $x$, we have to form a 9-digit number in which exactly one digit is repeated, and it is repeated exactly twice. So, one digit appears twice, and the remaining 7 digits appear once each.

First, select the digit that will be repeated: from $S$, this can be done in ${ }^9 C_1$ ways.

Now, from the remaining 8 digits, choose 7 digits that will appear once: this can be done in ${ }^8 C_7$ ways.

Now we have 9 digits in total, but one digit is repeated twice. So, the number of different arrangements is $=\frac{9!}{2!}$ ways.

$ x={ }^9 C_1 \times{ }^8 C_7 \times \frac{9!}{2!} $

2. Calculation for y

For $y$, we have to form a 9-digit number in which exactly two digits are repeated, and each of these is repeated exactly twice. So, two digits appear twice each (total 4 places), and the remaining 5 digits appear once each.

First, select the 2 digits that will be repeated from $S$: this can be done in ${ }^9 C_2$ ways.

Now, from the remaining 7 digits, choose 5 digits that will appear once: this can be done in ${ }^7 C_5$ ways.

Now we have 9 digits in total, with two digits repeated twice each. So, the number of different arrangements is $=\frac{9!}{2!2!}$ ways.

$ y={ }^9 C_2 \times{ }^7 C_5 \times \frac{9!}{2!2!} $

To find the relation between $x$ and $y$, divide $x$ by $y$:

To find the relation, divide x by y :

$ \frac{x}{y}=\frac{{ }^9 C_1 \times{ }^8 C_7 \times \frac{1}{2}}{{ }^9 C_2 \times{ }^7 C_5 \times \frac{1}{4}}=\frac{9 \times 8 \times \frac{1}{2}}{36 \times 21 \times \frac{1}{4}} $

So,

$\frac{x}{y}=\frac{36}{189}=\frac{4}{21}$

$\Rightarrow 21 x=4 y$

Arrang UDAYPUR in alphabetical order A, D, P, R, UU, Y

Number of words start with A. fix A and arrange DPRUUY $=\frac{6!}{2!}=\frac{6 \times 5 \times 4 \times 3 \times 2 \times 1}{2 \times 1}=360$ ways.

Similarly, number words starts from D, P, R will be same as there is arrangement of 6 letters in which $U$ is repeating twice.

So, total number of words starting with A, D, P \& R are $=4 \times 360=1440$.

Now, number of words starts with UA

fix UA and arrange D, P, R, U, Y = 5! ways = 120 .

the next letter is D after A, so number of words starts with UDAP arranging $(R, U, Y)=3!$ ways $=6$

So, number of words start with UDAR and UDAU is same $=3!+3!=12$

next is UDAY matches with pattern(UDAYPUR)

number of words start with UDAYP arrang(RU) $=2!=2$

number of words start with UDAYPR is UDAYPRU $=1$

final pattern UDAYPUR = 1

So, rank of word UDAYPUR is =

$ 1440+120+6+12+1+1=1580 . $

To find largest value of $n$ for which 60 ! is divisible by $40^n$

This means we need to find exponent of 40 in 60 !

$ 40=2^3 \times 5 $

Key Concept: formula for exponent of prime $p$ in $k!$ is given as $\left[\frac{k}{p}\right]+\left[\frac{k}{p^2}\right]+\left[\frac{k}{p^3}\right]+\ldots$ where [] is greatest integer function.

so, exponent of 2 in 60 !

$ \begin{aligned} & {\left[\frac{60}{2}\right]+\left[\frac{60}{2^2}\right]+\left[\frac{60}{2^3}\right]+\left[\frac{60}{2^4}\right]+\left[\frac{60}{2^5}\right]+\left[\frac{60}{2^6}\right]+\ldots} \\ & 30+15+7+3+1=56-\text { (1) } \end{aligned} $

exponent of 5 in 60 !

$ \begin{aligned} & {\left[\frac{60}{5}\right]+\left[\frac{60}{5^2}\right]+\left[\frac{60}{5^3}\right]+\ldots} \\ & 12+2+0 \cdots=14-(2) \end{aligned} $

Combining (1) and (2)

60 ! is divisible by $2^{56} \times 5^{14} \times N$, where $N$ is natural number.

60 ! is divisible by $4 \times\left(2^3\right)^{18} \times 5^{14} \times N$

60 ! is divisible by $4 \times\left(2^3\right)^4 \times\left(2^3\right)^{14} \times 5^{14} \times N$

$60!$ is divisible by $2^{14} \times 40^{14} \times N$

So largest value of $n$ is 14

To distribute 16 oranges to 4 children such that each child gets atleast one orange.

First distribute one orange to each child so that each child get atleast one.

Now, distribute remaining ( $\mathrm{n}=12$ ) oranges to ( $\mathrm{r}=4$ ) children using beggar's method $={ }^{n+r-1} C_{r-1}$

$ \begin{aligned} & ={ }^{12+4-1} C_{4-1}={ }^{15} C_3 \\\\ & =\frac{15 \times 14 \times 13}{3 \times 2 \times 1}=5 \times 7 \times 13=455 \end{aligned} $

$ \text { Exponent of } 7 \text { in } 101!=\left[\frac{101}{7}\right]+\left[\frac{101}{7^2}\right]+\ldots . . \text { (Where [.] is G.I.F.) }=14+12=16 $

Case (i) If $f(1)=2$ then ${ }^7 C_5$ functions

Case (ii) $f(1)=3$ then ${ }^6 C_5$

Case (iii) $f(1)=4$ then ${ }^5 C_5$

$ \Rightarrow 21+6+1=28 $

To count the number of distinct triangles:

Start with all possible triples of points.

From 12 points, the number of ways to choose any 3 is

$ \binom{12}{3}= \frac{12\cdot11\cdot10}{3\cdot2\cdot1}=220. $

Subtract the “invalid” triples that are collinear.

The only collinear sets of three points arise from the single line containing the 5 collinear points.

Number of ways to pick 3 points from those 5 is

$ \binom{5}{3}=10. $

Valid triangles = total triples − collinear triples.

$ 220-10 = 210. $

Hence, the total number of triangles that can be formed is 210.

Correct option: B

1 Captain, 1 vice-captain are already present

$\Rightarrow$ We need to select 8 players such that atleast 3 batsman and bowler must be there

Total = 155 ways

Total triangles

(2 points as $y=\frac{x}{2}, 1$ point on $y=\frac{x}{2}$)

$+2\left(\right.$ points and $y=\frac{x}{2}, 1$ point on $y=2 x$)

+(1 point on $y=2 x, 1)$ point on $y=\frac{x}{2}$ and origin)

$={ }^9 C_2,{ }^{12} C_1+{ }^9 C_1{ }^{12} C_2+{ }^9 C_1 \cdot{ }^{12} C_1 \cdot{ }^1 C_1$

$=1134$

Let $x, y, z$ be the number of box which are filled

$\Rightarrow 1 \leq x \leq 3,1 \leq y \leq 3,1 \leq z \leq 2$

Total ways $=(48)$ to fill boxes

Now to arrange $a, b, c, d$ and $e$

Number of ways will be 48.5! = 5760

To find the number of sequences of ten terms, each being either 0, 1, or 2, containing exactly five 1s and exactly three 2s, follow these steps:

Determine the Remaining Terms: Since there are 5 ones and 3 twos, you will need 2 zeros to fill the sequence (because $5 + 3 + 2 = 10$).

Calculate the Number of Arrangements: You have a total of 10 positions to fill with these numbers (5 ones, 3 twos, 2 zeros). The formula for calculating permutations of a multiset is:

$ \frac{10!}{5! \times 3! \times 2!} $

Where:

$10!$ is the factorial of the total number of terms.

$5!$ is the factorial for the number of 1s.

$3!$ is the factorial for the number of 2s.

$2!$ is the factorial for the number of 0s.

Result: Simplifying the calculation gives:

$ \frac{10!}{5! \times 3! \times 2!} = 2520 $

Thus, there are 2520 possible sequences with the given conditions.

(i) number of numbers created using

$1111133=\frac{7!}{5!2!} \Rightarrow 21$

(ii) number of numbers created using

$1111223=\frac{7!}{4!2!} \Rightarrow 105$

(iii) number of numbers created using

$\begin{aligned} & 1112222=\frac{7!}{4!3!} \Rightarrow 35 \\ & \text { Total }=161 \end{aligned}$

$\begin{aligned} & { }^n C_{r-1}=28,{ }^n C_r=56 \\ & \frac{{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}-1}}{{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}}=\frac{28}{56} \\ & \frac{\frac{n!}{(r-1)!(n-r+1)!}}{\frac{n!}{r!(n-r)!}}=\frac{1}{2} \\ & \frac{\mathrm{r}}{(\mathrm{n}-\mathrm{r}+1)}=\frac{1}{2} \\ & 3 \mathrm{r}=\mathrm{n}+1\quad\text{..... (i)} \end{aligned}$

$\begin{aligned} & \frac{{ }^n C_r}{{ }^n C_{r+1}}=\frac{56}{70} \\ & \frac{(r+1)}{(n-r)}=\frac{56}{70} \Rightarrow 9 r=4 n-5\quad\text{.... (ii)} \end{aligned}$

$\begin{aligned} & \text { By (i) & (ii) } \\ & \begin{array}{l} (r=3),(n=8) \\ A(4 \cos t, 4 \sin t) \quad B(2 \sin t,-2 \cos t) C\left(3 r-n, r^2-n-1\right) \\ A(4 \cos t, 4 \sin t) \quad B(2 \sin t,-2 \operatorname{cost}) C(1,0) \\ (3 x-1)^2+(3 y)^2=(4 \operatorname{cost}+2 \sin t)^2+(4 \sin t-\operatorname{cost})^2 \\ (3 x-1)^2+(3 y)^2=20 \quad \therefore \text { option }(1) \end{array} \end{aligned}$

Case I $5{ }_{---} 0$

Case II $5{ }_{---} 1$

$ \begin{array}{ll} 5 & 2 \\ 5 & 3 \\ 6 & 0 \\ 6 & 1 \\ 6 & 2 \\ 7 & 0 \end{array}$

Case IX $7{ }_{---} 1$

$9 \times(8 \times 8 \times 8)=4608$ but 50000 is not included, so total numbers $4608-1=4607$

$\begin{array}{ll} \text { C-I } & (3 \mathrm{G} \& 2 \mathrm{~B}) \&(1 \mathrm{G} \& 2 \mathrm{~B}) \\ \text { C-II } & (2 \mathrm{G} \& 3 \mathrm{~B}) \&(2 \mathrm{G} \& 1 \mathrm{~B}) \\ \text { C-III } & (1 \mathrm{G} \& 4 \mathrm{~B}) \&(3 \mathrm{G} \& 0 \mathrm{~B}) \end{array}$

$\begin{aligned} & \text { Total }=\text { C-I }+ \text { C-II }+ \text { C-III } \\ & ={ }^7 \mathrm{C}_2 \cdot{ }^3 \mathrm{C}_3 \cdot{ }^6 \mathrm{C}_2 \cdot{ }^5 \mathrm{C}_1+{ }^7 \mathrm{C}_3 \cdot{ }^3 \mathrm{C}_2 \cdot{ }^6 \mathrm{C}_1{ }^5 \mathrm{C}_2+{ }^7 \mathrm{C}_4 \cdot{ }^3 \mathrm{C}_1 \cdot{ }^6 \mathrm{C}_0 \cdot{ }^5 \mathrm{C}_3 \\ & =8925 \end{aligned}$

DAUGHTER

Total words $=8$ !

Total words in which vowels are together $=6!\times 3!$ words in which all vowels are not together

$\begin{aligned} & =8!-6!\times 3! \\ & =6![56-6] \\ & =720 \times 50 \\ & =36000 \end{aligned}$

Let's break the problem down step by step:

There are two blocks because all girls must stand together and all boys must stand together. The two blocks can be arranged in:

$2 \text{ ways} \quad \text{(i.e., girls first then boys, or boys first then girls).}$

The girls can be arranged among themselves in:

$3! = 6 \text{ ways.}$

For the boys (4 in total), they must be arranged such that the specific boys $B_1$ and $B_2$ are not adjacent.

First, calculate the total number of arrangements of 4 boys:

$4! = 24.$

Next, count the arrangements where $B_1$ and $B_2$ are adjacent. Think of $B_1$ and $B_2$ as a single unit. This unit can be arranged in:

$2! = 2 \text{ ways (since }B_1\text{ and }B_2\text{ can swap positions).}$

Now, with this new unit, we have 3 units in total (the $B_1B_2$ unit and the other 2 boys), which can be arranged in:

$3! = 6 \text{ ways.}$

So, the number of arrangements where $B_1$ and $B_2$ are adjacent is:

$2! \times 3! = 2 \times 6 = 12.$

Therefore, the number of valid arrangements for the boys where $B_1$ and $B_2$ are not adjacent is:

$24 - 12 = 12.$

Finally, multiply all the factors together:

$\text{Total ways} = 2 \times 6 \times 12 = 144.$

Thus, the number of ways in which the girls and boys can stand in the queue under the given conditions is $\boxed{144}.$

This corresponds to Option D.

First, note that we are choosing 5 distinct letters (in strictly increasing alphabetical order) such that the middle (third) letter is ‘M’. Symbolically, if we denote the chosen letters as:

$ L_1 < L_2 < L_3 < L_4 < L_5, $

we want $L_3 = \text{M}$. The English alphabet has 26 letters, and M is the $13^\text{th}$.

Step 1: Letters before M

The letters before M are $\{A, B, C, \ldots, L\}$.

There are 12 letters here ($A$ through $L$).

We need to pick 2 of these 12 letters to occupy $L_1$ and $L_2$.

The number of ways to choose 2 letters out of 12 is ${ }^{12} \mathrm{C}_2$.

Step 2: Letters after M

The letters after M are $\{N, O, P, \ldots, Z\}$.

There are 13 letters here ($N$ through $Z$).

We need to pick 2 of these 13 letters to occupy $L_4$ and $L_5$.

The number of ways to choose 2 letters out of 13 is ${ }^{13} \mathrm{C}_2$.

Step 3: Multiply the choices

Since these choices are independent (picking the two letters before M and two letters after M), the total number of ways is:

$ { }^{12} \mathrm{C}_2 \;\times\;{ }^{13} \mathrm{C}_2 $

Calculate each combination:

$ { }^{12} \mathrm{C}_2 = \frac{12 \times 11}{2} = 66, \quad { }^{13} \mathrm{C}_2 = \frac{13 \times 12}{2} = 78. $

So,

$ { }^{12} \mathrm{C}_2 \times { }^{13} \mathrm{C}_2 = 66 \times 78 = 5148. $

Answer: 5148 (Option B)

$\begin{aligned} & 2 M \\ & 2 A \\ & 2 T \\ & H, E, I, C, S \end{aligned}$

Case-I

2 Alike 2 Alike 1 Diff

${ }^3 C_2 \times{ }^6 C_1=18$

Case-II

2 Alike + 3 Diff

${ }^3 C_1 \times{ }^7 C_3=105$

Case-III

All different

${ }^8 C_5=56$

Total ways $=179$

$\begin{aligned} & A=\{2,3,5,7,11\} \\ & f(x)=\left[\log _2\left(x^2+\left[\frac{x^3}{5}\right]\right)\right] \\ & \text { Ranges } f(x)=\{2,3,5,6,8\} \end{aligned}$

$\text { Number of one-one } A \rightarrow R_f$

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

NAGPUR

Word at $315^{\text {th }}$ position

$\begin{aligned} & \text { A...... }=5!=120 \\ & \text { G....... }=5!=120 \\ & \text { NA..... }=4!=24 \\ & \text { NG..... }=4!=24 \\ & \text { NP..... }=4!=24 \end{aligned}$

..... Till 312 words

$313^{\text {th }} \text { word }=\text { NRAGPU }$

$314^{\text {th }}$ word $=$ $\mathrm{NRAGUP}$

$315^{\text {th }} \text { word }=\text { NRAPGU }$

Given $0 \leq r \leq n$. If $\binom{n+1}{r+1} : \binom{n}{r} : \binom{n-1}{r-1} = 55 : 35 : 21$, then we are to determine the value of $2n + 5r$.

Step-by-Step Solution:

1. Write the given proportions involving binomial coefficients:

$ \frac{n+1}{r+1} \times \binom{n}{r} : \binom{n}{r} : \frac{r}{n} \times \binom{n}{r} = 55 : 35 : 21 $

1. Simplify the proportions:

$ \frac{n+1}{r+1} = \frac{55}{35} \quad \text{and} \quad \frac{n}{r} = \frac{35}{21} $

1. From the simplified ratios, we establish the following two equations:

$ \frac{n+1}{r+1} = \frac{11}{7} \quad \Rightarrow \quad 7(n+1) = 11(r+1) \quad \Rightarrow \quad 7n - 11r = 4 \quad \text{.... (1)} $

$ \frac{n}{r} = \frac{5}{3} \quad \Rightarrow \quad 3n = 5r \quad \Rightarrow \quad 3n - 5r = 0 \quad \text{.... (2)} $

1. Solve equations (1) and (2) simultaneously:

• From equation (2), solve for $n$:

$ 3n = 5r \quad \Rightarrow \quad n = \frac{5r}{3} $

• Substitute $n$ into equation (1):

$ 7 \left(\frac{5r}{3}\right) - 11r = 4 \quad \Rightarrow \quad \frac{35r}{3} - 11r = 4 $

$ \frac{35r - 33r}{3} = 4 \quad \Rightarrow \quad \frac{2r}{3} = 4 \quad \Rightarrow \quad 2r = 12 \quad \Rightarrow \quad r = 6 $

• Substituting $r$ back to find $n$:

$ n = \frac{5 \times 6}{3} = 10 $

1. Compute $2n + 5r$:

$ 2n + 5r = 2 \times 10 + 5 \times 6 = 20 + 30 = 50 $

Thus, the value of $2n + 5r$ is 50.

To solve this problem, we need to determine the number of triangles formed by the vertices of a regular octagon such that none of the sides of the triangle is also a side of the octagon.

Let's start by counting the total number of triangles that can be formed using the 8 vertices of the octagon. The number of ways to choose 3 vertices out of 8 is given by the combination formula:

$ \binom{8}{3} = \frac{8!}{3!(8-3)!} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56 $

Next, we need to exclude those triangles that have at least one side coinciding with a side of the octagon. Let's analyze how many such invalid triangles there can be.

Consider each side of the octagon. For any given side, there are exactly 5 other vertices remaining (since we must exclude the two vertices that form the current side). Out of these 5 vertices, we can choose any 1 to form a triangle that has one side common with the octagon. Hence, for each side of the octagon, there are 5 such triangles.

Since the octagon has 8 sides, the total number of triangles that have at least one side as a side of the octagon is:

$ 8 \times 5 = 40 $

Therefore, the number of triangles whose sides do not coincide with any sides of the octagon is:

$ 56 - 40 = 16 $

So, the correct answer is:

Option B: 16

Given set $S=\left\{2^1, 2^2, \ldots 2^9\right\}$ which consist of 9 elements.

Maximum number of possible partitions (in set $A, B$ and $C$)

$={ }^9 C_3 \cdot{ }^6 C_3 \cdot{ }^3 C_3=1680$

To find the $50^{\text{th}}$ word formed by the letters of "BHBJO" as if listed in a dictionary, let's analyze the arrangement methodically. Given the letters are B, H, B, J, O, there are some repetitions with the letter B appearing twice.

First, calculate the total number of permutations of these letters:

$ \frac{5!}{2!} = 60 $

Let's arrange the letters in alphabetical order first: B, B, H, J, O.

We need to systematically count the words while following dictionary order:

1. Words starting with B:

• Next position letters: B, H, J, O
• Number of permutations: $\frac{4!}{1!} = 24$ words

Since 24 words starting with 'B' exist and are less than 50, Move to next starting letter alphabetically.

2. Words starting with H:

• Next position letters: B, B, J, O
• Number of permutations: $\frac{4!}{2!} = 12$ words

After 'B', tally becomes 24 (B-words) + 12 (H-words) = 36, needs more.

3. Words starting with J:

• Next position letters: B, B, H, O
• Number of permutations: $\frac{4!}{2!} = 12$ words

Now tally is 36 + 12 = 48 words.

Still 2 more to reach 50.

4. Words starting with O:

• Next position letters: B, B, H, J
• Number of permutations: $\frac{4!}{2!} = 12$ words
• Our answer must be here since 48 + 2 more = 50 total.

First permutation: $OBB H J$

Second permutation (50th word): $OBB J H$

So, the $50^{\text{th}}$ word is: OBBJH

Thus, the correct answer is Option A: OBBJH.

Number of points on side $A B=5$

Number of points on side $B C=6$

Number of points on side $A C=7$

Number of ways selecting three points from side

$A B={ }^5 C_3$

Number of ways selecting three points from side

$B C={ }^6 C_3$

Number of ways selecting three points from side

$A C={ }^7 C_3$

Total number of triangle possible formed using the points $P_1 P_2 \ldots P_{18}$

$\begin{aligned} & ={ }^{18} C_3-{ }^5 C_3-{ }^6 C_3-{ }^7 C_3 \\\\ & =816-10-20-35 \\\\ & =751 \end{aligned}$

To determine the number of ways in which five different employees can be seated in four indistinguishable offices, we will use the concept of partitioning of an integer.

The problem is equivalent to partitioning the number 5 (representing the 5 different employees) into at most 4 parts (representing the 4 indistinguishable offices) where each part represents the number of employees in each office.

The partitions of 5 into at most 4 parts are as follows :

1. $(5, 0, 0, 0)$ – One office has 5 employees, and the other three have none.


2. $(4, 1, 0, 0)$ – One office has 4 employees, another one has 1, and the other two have none.


3. $(3, 2, 0, 0)$ – One office has 3 employees, another one has 2, and the other two have none.


4. $(3, 1, 1, 0)$ – One office has 3 employees, two have 1 each, and one has none.


5. $(2, 2, 1, 0)$ – Two offices have 2 employees each, one has 1, and one has none.


6. $(2, 1, 1, 1)$ – One office has 2 employees, and the other three have 1 each.

However, since the offices are indistinguishable, we must not count partitions that differ only by the order of the parts. This means that partitions like $(5, 0, 0, 0)$, $(0, 5, 0, 0)$, $(0, 0, 5, 0)$, and $(0, 0, 0, 5)$ all represent the same scenario and thus are counted as one.

So we end up with 6 distinct partition scenarios.

Now, for each partition, we need to find the number of ways to assign the employees according to each partition :

1. $(5, 0, 0, 0)$ – All employees are in one office. There is 1 way to do this because the offices are indistinguishable.


2. $(4, 1, 0, 0)$ - For the group with 4 employees, there are ${}^5{C_4}$ ways to choose which 4 out of the 5 will be together. Then the remaining employee is automatically assigned to the second office. So, there are ${}^5{C_4}$ ways for this partition.


3. $(3, 2, 0, 0)$ - For the group with 3 employees, there are ${}^5{C_3}$ ways to choose them, and for the group with 2 employees, there are ${}^2{C_2}$ ways to choose them (which is essentially 1 way since the last 2 employees are automatically grouped). But the 2 offices these groups can occupy are indistinguishable, so we do not multiply by 2. So, there are ${}^5{C_3}$ ways for this partition.


4. $(3, 1, 1, 0)$ - For the group with 3 employees, there are ${}^5{C_3}$ ways. Then we have two indistinguishable offices, each with 1 employee. The order does not matter as offices are indistinguishable. So, there are simply ${}^5{C_3}$ ways.


5. $(2, 2, 1, 0)$ – For the first group of 2 employees, there are ${}^5{C_2}$ ways to choose them. For the next group of 2 employees, there are ${}^3{C_2}$ ways from the remaining 3. Then, the last employee is alone, and since the offices are indistinguishable, there are no further combinations to consider. However, we have double-counted since the two groups of two are indistinguishable. To adjust for this, we divide by 2. This gives us $\frac{{}^5{C_2} \cdot {}^3{C_2}}{2}$ ways.


6. $(2, 1, 1, 1)$ - For the group with 2 employees, there are ${}^5{C_2}$ ways to choose them. The other three employees are each in their own office, with no further combinations since the offices are indistinguishable, so there are ${}^5{C_2}$ ways for this partition.

Let's compute these cases :

• $(5, 0, 0, 0)$ corresponds to $1$ way.


• $(4, 1, 0, 0)$ corresponds to ${}^5{C_4} = 5$ ways.


• $(3, 2, 0, 0)$ corresponds to ${}^5{C_3} = 10$ ways.


• $(3, 1, 1, 0)$ corresponds to ${}^5{C_3} = 10$ ways.


• $(2, 2, 1, 0)$ corresponds to $\frac{{}^5{C_2} \cdot {}^3{C_2}}{2} = \frac{10 \cdot 3}{2} = 15$ ways.


• $(2, 1, 1, 1)$ corresponds to ${}^5{C_2} = 10$ ways.

Adding up all the ways we get :

$ 1 + 5 + 10 + 10 + 15 + 10 = 51 $

Therefore, the number of ways five different employees can sit into four indistinguishable offices ($\mathrm{n}$) is 51, which corresponds to Option C.

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

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

The letters of "MONDAY" arranged in alphabetical order are: A, D, M, N, O, Y.

1. Fix A as the first letter, we can arrange the remaining 5 letters in 5! ways = 120 ways.
2. Fix D as the first letter, we can arrange the remaining 5 letters in 5! ways = 120 ways.

So far, we have 120 (for A) + 120 (for D) = 240 words.

Now, let's proceed with words starting with M:

1. Fix MA as the first two letters, we can arrange the remaining 4 letters in 4! ways = 24 ways.
2. Fix MD as the first two letters, we can arrange the remaining 4 letters in 4! ways = 24 ways.
3. Fix MN as the first two letters, we can arrange the remaining 4 letters in 4! ways = 24 ways.

Now, we have 240 (for A and D) + 24 (for MA) + 24 (for MD) + 24 (for MN) = 312 words.

After MN, we consider words that start with MO:

1. Fix MOA as the first three letters, we can arrange the remaining 3 letters in 3! ways = 6 ways.
2. Fix MOD as the first three letters, we can arrange the remaining 3 letters in 3! ways = 6 ways.

Adding these to the total, we have 312 (previous total) + 6 (for MOA) + 6 (for MOD) = 324 words.

Next, we consider words that start with 'MON'. The word 'MONDAY' comes after 'MONAD' in dictionary order, so:

1. Fix 'MONA' as the first four letters. The remaining 2 letters ('D' and 'Y') can be arranged in 2! ways, which gives us 2 more words: 'MONADY' and 'MONAYD'.

So, we have 324 (previous total) + 2 (for 'MONADY' and 'MONAYD') = 326 words.

Finally, we have the word 'MONDAY' itself, which is the 327th word.

First, let's determine the number of elements in the Cartesian product $A \times B$. If set $A$ has 5 elements and set $B$ has 2 elements, then the number of elements in $A \times B$ is:

$|A \times B| = |A| \times |B| = 5 \times 2 = 10$

We need to find the number of subsets of $A \times B$ with at least 3 elements and at most 6 elements. The total number of elements in $A \times B$ is 10, so we must consider the subsets containing 3, 4, 5, and 6 elements.

The number of ways to choose $r$ elements from a set of 10 elements is given by the binomial coefficient:

$\binom{10}{r} = \frac{10!}{r!(10-r)!}$

We will find the sum of the binomial coefficients for $r = 3, 4, 5, 6$:

First, $\binom{10}{3}$:

$\binom{10}{3} = \frac{10!}{3!7!} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120$

Next, $\binom{10}{4}$:

$\binom{10}{4} = \frac{10!}{4!6!} = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = 210$

Next, $\binom{10}{5}$:

$\binom{10}{5} = \frac{10!}{5!5!} = \frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2 \times 1} = 252$

Next, $\binom{10}{6}$:

$\binom{10}{6} = \frac{10!}{6!4!} = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = 210$

Finally, we sum these values:

$\binom{10}{3} + \binom{10}{4} + \binom{10}{5} + \binom{10}{6} = 120 + 210 + 252 + 210 = 792$

Therefore, the number of subsets of $A \times B$ each having at least 3 and at most 6 elements is:

Option D - 792

$a+b=23\lambda+2$

$\lambda=0,1,2,$ ...., but $\lambda$ cannot be even as $a+b$ is odd

$\lambda=1$ $(a, b)\to12$ pairs

$\lambda=3$ $(a,b)\to35$ pairs

$\lambda=5$ $(a,b)\to42$ pairs

$\lambda=7$ $(a,b)\to19$ pairs

$\lambda=9$ $(a,b)\to0$ pairs

$ \vdots $

Total $=12+35+42+19=108$

Lets arrange the letters of OUGHT in alphabetical order.

G, H, O, T, U

Number divisible by 3 = 300

Number divisible by 4 = 225

Number divisible by 12 = 75

Number divisible by 48 = 18

Total required number = 300 + 225$-$ 75 $-$ 18 = 432

$\therefore$ Option (1) is correct.