Permutations and Combinations

Given:

• The lift goes up to the $10^{\text {th }}$ floor.

• The lift does not stop at $1^{\text {st }}, 2^{\text {nd }}, 3^{\text {rd }}$ floors.

• So, the persons can get down only at the remaining floors.

$ 4,5,6,7,8,9,10 \Rightarrow 7 \text { floors } $

• Now, all three persons must exit at three different floors (no two can get down at the same floor).

• Also, the three persons are distinct, so who gets down where matters.

First, choose which 3 floors (out of 7) will be used:

$ { }^7 C_3=\frac{7 \cdot 6 \cdot 5}{{3} \cdot 2 \cdot 1}=35 $

Next, assign (arrange) the 3 distinct persons to these 3 chosen floors:

Number of arrangements of 3 persons:

$ 3!=6 $

So, total number of ways:

$ \text { Total ways }={ }^7 C_3 \times 3!=35 \times 6=210 $

To find numbers of number greater than 5000 and less than 9000 so number will start with 5 or 9 of the form 5 abc and $a, b, c \in\{0,1,2,5,9\}$ also the number is divisible by 3 .

This means $5+\mathrm{a}+\mathrm{b}+\mathrm{c}$ is divisible by 3

so the sum $\mathrm{a}+\mathrm{b}+\mathrm{c}$ is of the type $3 \mathrm{n}+1$.

unit place c determined by the sum of previous 3 places.

rem(a $+b)=$ remainder of $a+b$

rem(a+b)	combination (a,b)	value of c	total = combination (a, b) × c(choice)
0	(0, 0), (0, 9), (9, 0), (9, 9), (1, 2), (2, 1), (5, 1), (1, 5)	1	8 × 1 = 8
1	(0, 1), (9, 1), (1, 0), (1, 9), (2, 2), (2, 5), (5, 2), (5, 5)	0, 9	8 × 2 = 16
2	(0, 2), (0, 5), (9, 2), (9, 5), (2, 0), (2, 9), (5, 0), (5, 9), (1, 1)	2, 5	9 × 2 = 18

Total = $8+16+18=42$

for set $S$ : 4-digit numbers $a b c d$ with $a>b>c>d$

digits are chosen from $\{0,1,2, \ldots, 9\}$

since order is strictly fixed, any 4 distinct digits chosen will form exactly one such number. number of ways to choose 4 digits from 10 is ${ }^{10} C_4$

$ n(S)=\frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1}=210 $

for set $P$ : 5-digit numbers with product of digits $=20$

possible sets of digits (using digits 1-9):

case 1: $\{5,4,1,1,1\}$

number of arrangements $=\frac{5!}{3!}=20$

case 2: $\{5,2,2,1,1\}$

number of arrangements $=\frac{5!}{2!2!}=30$

$ n(P)=20+30=50 $

calculating final value.

$ n(S)+n(P)=210+50=260 $

the value of $n(S)+n(P)$ is 260 .

Letter frequency

$ \mathrm{P}: 3 $

$\mathrm{R}, \mathrm{Q}: 2$

$S, T, U, V: 1$

4 letter words can be of type

$\Rightarrow \mathrm{ABCD}$ or $\mathrm{AABC}, \mathrm{AABB}, \mathrm{AAAB}$

$ \begin{aligned} & \Rightarrow{ }^7 C_4 \cdot 4!+\left({ }^3 C_1\right) \cdot{ }^6 C_2 \cdot \frac{4!}{2!}+{ }^3 C_2 \cdot \frac{4!}{2!1!}+\left({ }^1 C_1\right) \cdot{ }^6 C_1 \cdot \frac{4!}{3!} \\ & =1422 \end{aligned} $

$ { }^4 C_2 \cdot{ }^5 C_2 \cdot{ }^4 C_1+{ }^4 C_2 \cdot{ }^5 C_2 \cdot{ }^4 C_1+{ }^4 C_2 \cdot{ }^4 C_2 \cdot{ }^5 C_1=240+240+180=660 $

$ \begin{aligned} & 5 \mid 6^m+9^n \\ & \Rightarrow 5 \mid 1^m+(-1)^n \end{aligned} $

$\Rightarrow \quad m$ and $n$ has to be opposite parity.

$ { }^2 C_1 \times{ }^{25} C_1 \cdot{ }^{25} C_1=625 \times 2=1250 $

For, $m+n=K^2$ for some prime $K$.

$ \begin{aligned} & m+n \in\{2,3, \ldots, 100\} \\ & m+n=4,9,25,49 \end{aligned} $

$m+n=4 \Rightarrow 3$ pairs

$m+n=9 \Rightarrow 8$ pairs

$m+n=25 \Rightarrow 24$ pairs

$m+n=49 \Rightarrow 48$ pairs

$\Rightarrow 83$ pairs

$\Rightarrow 1333$

An equivalence relation divides the given set into disjoint groups, called equivalence classes. Together, these classes form a complete partition of the set.

Here, the set is $ S = \{1, 2, 3, \ldots, 10\} $. So, we must divide it into parts (subsets) such that the total number of elements across all parts is 10.

For an equivalence relation, if one class has $ n_i $ elements, then all possible ordered pairs inside this class are related. Hence, it contributes $ n_i^2 $ related pairs to the relation.

Because the total number of related pairs is given as 42, we need to find positive integers $ n_1, n_2, \ldots, n_k $ satisfying both:

$ n_1 + n_2 + \ldots + n_k = 10 $

and

$ n_1^2 + n_2^2 + \ldots + n_k^2 = 42 $

Now, we search for combinations of squares that add up to 42, while the corresponding numbers add up to 10.

Possible combinations are:

$ 42 = 36 + 4 + 1 + 1 \ \Rightarrow \ \{6, 2, 1, 1\} $

$ 42 = 25 + 16 + 1 \ \Rightarrow \ \{5, 4, 1\} $

Hence, the partitions can be $ \{6, 2, 1, 1\} $ and $ \{5, 4, 1\} $.

Case 1: Partition {5, 4, 1}

First choose 5 elements out of 10 for one class, then choose 4 out of the remaining 5 for the next class, and the last one forms a single-element class. The number of ways is:

$ {^{10}C_5} \times {^{5}C_4} \times {^{1}C_1} = 1260 $

Case 2: Partition {6, 2, 1, 1}

First choose 6 elements out of 10, then 2 from the remaining 4, and the last two each become single-element classes. Since the two single-element groups are identical, we divide by 2! to adjust for overcounting. So the number of ways is:

$ {^{10}C_6} \times {^{4}C_2} \times \frac{{^{2}C_1 \times ^{1}C_1}}{2} = 1260 $

Therefore, the total number of such equivalence relations is the sum of both cases:

$ 1260 + 1260 = 2520 $

We have to distribute 10 identical red pens and 14 identical blue pens among four persons so that each person gets 6 pens in total.

For every possible way of giving out the red pens, the blue pens will automatically be distributed, because the total number of pens each person receives is fixed at 6. Hence, we only need to count the number of ways to distribute the red pens.

Let the number of red pens received by the four persons be $ x_1, x_2, x_3, x_4 $. Then,

$ x_1 + x_2 + x_3 + x_4 = 10 $

and since each person can receive a maximum of 6 pens, $ 0 \leq x_i \leq 6 $.

The number of non-negative integer solutions to the equation $ x_1+x_2+x_3+x_4=10 $ with $ x_i \le 6 $ can be obtained by finding the coefficient of $ x^{10} $ in the expansion of $ (1+x+x^2+\ldots+x^6)^4 $.

We know that:

$ (1+x+x^2+\ldots+x^6) = \frac{1-x^7}{1-x} $

Therefore,

$ (1+x+x^2+\ldots+x^6)^4 = \left( \frac{1-x^7}{1-x} \right)^4 = (1-x^7)^4(1-x)^{-4} $

Now, we need the coefficient of $ x^{10} $ in this expansion. Expanding $ (1-x^7)^4 $, we get:

$ (1-x^7)^4 = 1 - 4x^7 + \ldots $

Thus, the required coefficient of $ x^{10} $ is:

$ \text{Coeff. of } x^{10} \text{ in } (1-x)^{-4} - 4 \times \text{Coeff. of } x^{3} \text{ in } (1-x)^{-4} $

We know that the coefficient of $ x^r $ in $ (1-x)^{-n} $ is $ {^{n+r-1}C_{r}} $. Hence,

$ {^{13}C_3} - 4 \times {^{6}C_3} = 286 - 80 = 206 $

Therefore, the total number of ways = 206.

For player A to win the series, A must win the $5^\text{th}$ game before B does.

So, when the series ends:

• A has won exactly $5$ games.

• B can have won $0,1,2,3,$ or $4$ games.

Now count each case.

Case 1: B wins $0$ games

Then the sequence is only

$ AAAAA $

Number of ways $= 1$

Case 2: B wins $1$ game

The series has $6$ games, and the last game must be won by A.

So in the first $5$ games, A must win $4$ games and B must win $1$ game.

Number of ways

$ \binom{5}{1} = 5 $

Case 3: B wins $2$ games

The series has $7$ games, and the last game must be won by A.

So in the first $6$ games, A must win $4$ games and B must win $2$ games.

Number of ways

$ \binom{6}{2} = 15 $

Case 4: B wins $3$ games

The series has $8$ games, and the last game must be won by A.

So in the first $7$ games, A must win $4$ games and B must win $3$ games.

Number of ways

$ \binom{7}{3} = 35 $

Case 5: B wins $4$ games

The series has $9$ games, and the last game must be won by A.

So in the first $8$ games, A must win $4$ games and B must win $4$ games.

Number of ways

$ \binom{8}{4} = 70 $

Total number of ways

$ 1 + 5 + 15 + 35 + 70 = 126 $

Hence, the number of ways in which player A wins the series is

$ \boxed{126} $

$\begin{aligned} & m=6 a, n=6 b \\ & \text { So } \operatorname{gcd}(m, n)=6 \Rightarrow \operatorname{gcd}(a, b)=1 \\ & m=6 a \geq 10 \Rightarrow a \geq\left[\frac{10}{6}\right]=2 \\ & m=6 a \leq 99 \Rightarrow a \leq\left[\frac{99}{6}\right]=16 \end{aligned}$

So $a, b \in\{2,3, \ldots, 16\}$, and we count how many coprime pairs $(a, b)$ with $a< b, \operatorname{gcd}(a, b)=1$

$\begin{aligned} & a=2 \Rightarrow b=3,5,7,9,11,13,15 \Rightarrow 7 \\ & a=3 \Rightarrow \mathrm{~b}=4,5,7,8,10,11,13,14,16 \Rightarrow 9 \\ & a=4 \Rightarrow b=5,7,9,11,13,15 \Rightarrow 6 \\ & a=5 \Rightarrow \mathrm{~b}=6,7,8,9,11,12,13,14,16 \Rightarrow 9 \\ & a=6 \Rightarrow \mathrm{~b}=7,11,13 \Rightarrow 3 \\ & a=7 \Rightarrow b=8,9,10,11,12,13,15,16 \Rightarrow 8 \\ & a=8 \Rightarrow \mathrm{~b}=9,11,13,15 \Rightarrow 4 \\ & a=9 \Rightarrow \mathrm{~b}=10,11,13,14,16 \Rightarrow 5 \\ & a=10 \Rightarrow b=11,13 \Rightarrow 2 \\ & a=11 \Rightarrow b=12,13,14,15,16 \Rightarrow 5 \\ & a=12 \Rightarrow b=13,17 \times \rightarrow \text { only } 13 \text { is valid } \Rightarrow 1 \\ & a=13 \Rightarrow b=14,15,16 \Rightarrow 3 \\ & a=14 \Rightarrow b=15, \Rightarrow 1 \\ & a=15 \Rightarrow b=16 \Rightarrow 1 \end{aligned}$

$\begin{aligned} & \text { Total }=7+9+6+9+3+8+4+5+2+5+1 \\ & +3+1+1=64\end{aligned}$

Firstly, by this rule, we note:

$ P(W_1) = p $

$ P(W_2) = 2p $

$ P(W_3) = 4p $

…

$ P(W_n) = 2^{n-1}p $

To find the initial probability $ p $, consider the total probability must sum to 1 across all 120 possible words (since $ 5! = 120 $):

$ \sum_{n=1}^{120} P(W_n) = 1 $

This is a geometric series sum where:

$ p(1 + 2 + 2^2 + \ldots + 2^{119}) = 1 $

Since the series sum $ 1 + 2 + 2^2 + \ldots + 2^{119} $ is equal to $ 2^{120} - 1 $, we have:

$ p(2^{120} - 1) = 1 \Rightarrow p = \frac{1}{2^{120} - 1} $

Thus, the probability for the $ n $-th word is:

$ P(W_n) = \frac{2^{n-1}}{2^{120} - 1} \quad \text{(i)} $

Next, determine the position of "CDBEA". Starting from the first letter:

Words beginning with 'A': $ 4! = 24 $

Words beginning with 'B': $ 4! = 24 $

Words beginning with 'C':

CA*: $ 3! = 6 $

CB*: $ 3! = 6 $

CDA$ **: 2! = 2 $

CDBA*: $ 1! = 1 $

Summing these, the position of "CDBEA" is the 64th word.

Substitute into equation (i):

$ P(\text{CDBEA}) = P(W_{64}) = \frac{2^{63}}{2^{120} - 1} $

Given $ P(\text{CDBEA}) = \frac{2^\alpha}{2^\beta - 1} $, we find:

$ \alpha = 63 $

$ \beta = 120 $

Thus, the sum $ \alpha + \beta = 63 + 120 = 183 $.

When numbers are uniformly distributed, half of them have even digit sums and half have odd digits sums.

Number of 7-digit numbers with even digit sum =

$\frac{1}{2} \cdot 9 \cdot 10^6=4.5 \cdot 10^6$

Note that $9 \cdot 5 \cdot 10^5=4.5 \cdot 10^6$

$m+n=9+5=14$

(i) Single letter is used, then no. of words $=5$

(ii) Two distinct letters are used, then no. of words

${ }^5 \mathrm{C}_2 \times\left(\frac{6!}{2!4!} \times 2+\frac{6!}{3!3!}\right)=10(30+20)=500$

(iii) Three distinct letters are used, then no. of words

${ }^5 \mathrm{C}_3 \times \frac{6!}{2!2!2!}=900$

Total no. of words $=1405$

$\begin{array}{|c|c|c|} \hline \mathrm{x} & \mathrm{y} & \mathrm{z} \\ \hline \end{array}$

Let $\mathrm{x}=2 \Rightarrow \mathrm{y}+\mathrm{z}=13$

$(4,9),(5,8),(6,7),(7,6),(8,5),(9,4), \rightarrow 6$

Let $x=3 \rightarrow y+z=12$

$(3,9),(4,8), \ldots \ldots . .,(9,3) \rightarrow 7$

Let $x=4 \rightarrow y+z=11$

$(2,9),(3,8), \ldots \ldots \ldots,(9,1) \rightarrow 9$

Let $x=5 \rightarrow y+z=10$

$(1,9),(2,8), \ldots \ldots . .,(9,1) \rightarrow 10$

Let $x=6 \rightarrow y+z=9$

$(0,9),(1,8), \ldots \ldots . .,(9,0) \rightarrow 9$

Let $\mathrm{x}=7 \rightarrow \mathrm{y}+\mathrm{z}=8$

$(0,9),(1,7), \ldots \ldots . .,(8,0) \rightarrow 9$

Let $x=8 \rightarrow y+z=7$

$(0,7),(1,6), \ldots \ldots . .,(7,0) \rightarrow 8$

Let $x=9 \rightarrow y+z=6$

$(0,6),(1,5), \ldots \ldots \ldots,(6,0) \rightarrow 7$

Total $=6=7+8+9+10+9+8+7=64$

392 Ans.

No, of 3 digits $=999-99=900$

No. of 3 digit numbers divisible by 2 & 3 i.e. by 6

$\frac{900}{6}=150$

No. of 3 digit numbers divisible by 4 & 9 i.e. by 36

$\frac{900}{36}=25$

$\therefore$ No of 3 digit numbers divisible by $2 \& 3$ but not by $4 \& 9$

$150-25=125$

A : number of ways that all boys sit together $=5!\times 5!$

B : number of ways if no 2 boys

sit together $=4!\times 5!$

$\mathrm{A} \cap \mathrm{B}=\phi$

Required no. of ways $=5!\times 5!+4!\times 5!=17280$

Problem restated: Let $R$ be a relation on $X = \{a,b,c,d,e,f\}$.

We want $R$ to be reflexive and symmetric, and contain exactly 10 elements.

Let $\mathcal{S} = \{R : R \text{ satisfies this}\}$. Find $|\mathcal{S}|$.

Step 1. Size of the universe

$|X| = 6$.

So $X \times X$ has $36$ ordered pairs.

Step 2. Reflexivity

Reflexive means we must include all $(x,x)$ pairs.

So there are 6 forced elements.

So any reflexive relation on $X$ automatically contains:

$ \{(a,a),(b,b),(c,c),(d,d),(e,e),(f,f)\}. $

So our reflexive relation starts with 6 elements.

Step 3. Symmetry

For off-diagonal elements:

If we include $(x,y)$ (for $x \neq y$), we must also include $(y,x)$.

So these off-diagonal pairs come in symmetric pairs (each of size 2).

There are $\binom{6}{2} = 15$ distinct unordered pairs $\{x,y\}$, each would correspond to either including both $(x,y)$ and $(y,x)$, or excluding both. So think of them as independent choices.

Step 4. Required size

Total size should be 10.

We already have 6 diagonal pairs.

So we need 4 more pairs.

But each symmetric pair contributes 2 elements.

So to get 4 additional, we must include exactly 2 symmetric pairs.

Step 5. Counting

Number of unordered pairs = 15.

We must choose 2 of them to include.

So total = $\binom{15}{2} = 105$.

Final Answer:

$ |\mathcal{S}| = \boxed{105} $

Number in this range will be 3-digit number.

$N=\overline{a b c}$ such that $a+b+c=14$

Also, $a \geq 1, \quad a, b, c \in\{0,1,2, \ldots 9\}$

Case I

All 3-digit same

$\Rightarrow 3 a=14$ not possible

Case II

Exactly 2 digit same:

$\Rightarrow 2 a+c=14$

$\begin{aligned} & (a, c) \in\{(3,8),(4,6),(5,4),(6,2),(7,0)\} \\ & \Rightarrow\left(\frac{3!}{2!}\right) \text { ways } \Rightarrow 5 \times 3-1 \\ & =15-1=14 \end{aligned}$

Case III

All digits are distinct

$a+b+c=14$

without losing generality $a > b > c$

$\begin{aligned} & (a, b, c) \in\left\{\begin{array}{l} (9,5,0),(9,4,1),(9,3,2) \\ (8,6,0),(8,5,1),(8,4,2) \\ (7,6,1),(7,5,2),(7,4,3) \\ (6,5,3) \end{array}\right. \\ & \Rightarrow 8 \times 3!+2(3!-2!)=48+8=56 \\ & =0+14+56=70 \end{aligned}$

To solve this problem, we need to find the number of 3-digit numbers formed using the digits 2, 3, 4, 5, and 7, with no repetition of digits allowed, and these numbers should not be divisible by 3. Let's break down the solution step-by-step:

1. Calculating the total number of 3-digit numbers without repetition:

The number of ways to form a 3-digit number from 5 unique digits (2, 3, 4, 5, 7) without repetition can be calculated using permutations:

The total number of permutations for choosing 3 digits out of 5 and arranging them is given by:

$5P3 = \frac{5!}{(5-3)!} = \frac{5 \times 4 \times 3 \times 2 \times 1}{2 \times 1} = 5 \times 4 \times 3 = 60$

So, there are 60 possible 3-digit numbers that can be formed from the digits 2, 3, 4, 5, and 7 without repetition.

2. Finding the 3-digit numbers divisible by 3:

A number is divisible by 3 if the sum of its digits is divisible by 3. Let's consider the sums of every combination of these three digits to find out which sums are divisible by 3:

Possible sums of combinations:

• 2 + 3 + 4 = 9 (divisible by 3)
• 2 + 4 + 7 = 13
• 2 + 5 + 7 = 14
• 2 + 3 + 5 = 10
• 2 + 3 + 7 = 12 (divisible by 3)
• 3 + 4 + 5 = 12 (divisible by 3)
• 3 + 4 + 7 = 14
• 3 + 5 + 7 = 15 (divisible by 3)
• 4 + 5 + 7 = 16

The combinations whose sums are divisible by 3 are:

• 2, 3, 4
• 2, 3, 7
• 3, 4, 5
• 3, 5, 7

Since the sum of the digits is divisible by 3 for these combinations, any permutation of these sets will yield a number divisible by 3:

The number of 3-digit numbers that can be formed from each set of 3 digits is:

$3! = 6$

So, the total number of 3-digit numbers divisible by 3 is:

$4 \text{ sets} \times 6 \text{ permutations per set} = 24$

3. Calculating the 3-digit numbers not divisible by 3:

To find the 3-digit numbers not divisible by 3, we subtract the number of those divisible by 3 from the total number of 3-digit numbers:

$60 - 24 = 36$

Therefore, the number of 3-digit numbers that can be formed using the digits 2, 3, 4, 5, and 7 without repetition, and which are not divisible by 3, is equal to 36.

Number of ways $=$ coefficient of $x^{16}$ in $\left(x+x^2+\ldots+\right.$ $\left.x^6\right)^4$

$=$ coefficient of $x^{16}$ in $\left(1-x^6\right)^4(1-x)^{-4}$

$=$ coefficient of $x^{16}$ in $\left(1-4 x^6+6 x^{12} \ldots\right)(1-x)^{-4}$

$={ }^{15} C_3-4 \cdot{ }^9 C_3+6=125$

Group A	Group B	Ways
$4m$ $3m+1w$ $2m+2w$ $1m+3w$ $4w$	$4w$ $1m+3w$ $2m+2w$ $3m+w$ $4m$	${ }^4 C_4 \cdot{ }^4 C_4 \quad=1$ ${ }^4 C_1 \cdot{ }^5 C_1 \cdot{ }^5 C_1^4 C_3 \quad=400$ ${ }^4 C_2 \cdot{ }^5 C_2{ }^5 C_2{ }^4 C_2 \quad=3600$ ${ }^4 C_1{ }^5 C_3{ }^5 C_3{ }^4 C_1 \quad=1600$ ${ }^5 C_4{ }^5 C_4 \quad=25$

$\begin{aligned} \text { Total ways } & =1+400+3600+1600+25 \\ & =5626 \end{aligned}$

$ x+2 y+3 z=42 $

$ x, y, z \geq 0 $

as

$\begin{array}{ll}z=0 & x+2 y=42 \Rightarrow 22 \text { cases } \\\\ z=1 & x+2 y=39 \Rightarrow 20 \text { cases } \\\\ z=2 & x+2 y=36 \Rightarrow 19 \text { cases } \\\\ z=3 & x+2 y=33 \Rightarrow 17 \text { cases } \\\\ z=4 & x+2 y=30 \Rightarrow 16 \text { cases } \\\\ z=5 & x+2 y=27 \Rightarrow 14 \text { cases }\end{array}$

$z=6 \quad x+2 y=24 \Rightarrow 13$ cases

$z=7 \quad x+2 y=21 \Rightarrow 11$ cases

$z=8 \quad x+2 y=18 \Rightarrow 10$ cases

$z=9 \quad x+2 y=15 \Rightarrow 8$ cases

$z=10 x+2 y=12 \Rightarrow 7$ cases

$z=11 x+2 y=9 \Rightarrow 5$ cases

$z=12 x+2 y=6 \Rightarrow 4$ cases

$z=13 x+2 y=3 \Rightarrow 2$ cases

$z=14 x+2 y=0 \Rightarrow 1$ case

Total = 169

We have III, TT, D, S, R, B, U, O, N

Number of words with selection (a, a, a, b)

$={ }^8 \mathrm{C}_1 \times \frac{4 !}{3 !}=32$

Number of words with selection (a, a, b, b)

$=\frac{4 !}{2 ! 2 !}=6$

Number of words with selection (a, a, b, c)

$={ }^2 \mathrm{C}_1 \times{ }^8 \mathrm{C}_2 \times \frac{4 !}{2 !}=672$

Number of words with selection (a, b, c, d)

$\begin{aligned} & ={ }^9 \mathrm{C}_4 \times 4 !=3024 \\\\ & \therefore \text {Total }=3024+672+6+32 \\\\ & =3734 \end{aligned}$

The problem involves choosing 15 questions out of a total of 20 available questions, with the constraint that at least 4 questions must be chosen from each of the three sections A, B, and C. To evaluate the total number of ways a student can select these questions, we need to consider every possible combination of questions from sections A, B, and C that sum up to 15 questions while respecting the constraints.

Section A has 8 questions, Section B and C each have 6 questions. The student must choose at least 4 questions from each section, which satisfies the minimum requirement. However, since the student is to attempt a total of 15 questions, there are several combinations to consider, as outlined below:

• Choosing 4 questions from A, 5 from B, and 6 from C
• Choosing 4 questions from A, 6 from B, and 5 from C
• Choosing 7 questions from A, 4 from B, and 4 from C
• Choosing 6 questions from A, 5 from B, and 4 from C
• Choosing 6 questions from A, 4 from B, and 5 from C
• Choosing 5 questions from A, 5 from B, and 5 from C
• Choosing 5 questions from A, 6 from B, and 4 from C
• Choosing 5 questions from A, 4 from B, and 6 from C

$ \begin{array}{|c|c|c|l|c|} \hline \mathrm{A} & \mathrm{B} & \mathrm{C} & \Rightarrow & \begin{array}{c} \text { No. of } \\ \text { question } \end{array} \\ \hline 4 & 5 & 6 & \rightarrow & { }^8 C_4{ }^6 C_5{ }^6 C_6 \\ \hline 4 & 6 & 5 & \rightarrow & { }^8 C_4{ }^6 C_6{ }^6 C_5 \\ \hline 7 & 4 & 4 & \rightarrow & { }^8 C_7{ }^6 C_4{ }^6 C_4 \\ \hline 6 & 5 & 4 & \rightarrow & { }^8 C_6{ }^6 C_5{ }^6 C_4 \\ \hline 6 & 4 & 5 & \rightarrow & { }^8 C_6{ }^6 C_4{ }^6 C_5 \\ \hline 5 & 5 & 5 & \rightarrow & { }^8 C_5{ }^6 C_5{ }^6 C_5 \\ \hline 5 & 6 & 4 & \rightarrow & { }^8 C_5{ }^6 C_6{ }^6 C_4 \\ \hline 5 & 4 & 6 & \rightarrow & { }^8 C_5{ }^6 C_4{ }^6 C_6 \\ \hline \end{array} $

Total ways of select $=11376$

Words starting with $\mathrm{E}=360$

Words starting with $\mathrm{GE=60}$

Words starting with $\mathrm{GN=60}$

Words starting with $\mathrm{GTE=24}$

Words starting with $\mathrm{GTN=24}$

Words starting with $\mathrm{GTT=24}$

GTWENTY $=1$

Total $=553$

$|a-b| \geq 2 \text { or }|b-a|=2$

Total

$\begin{array}{lll} a=1 & b=3,4,5,6 & 8 \\ a=2 & b=4,5,6 & 6 \\ a=3 & b=5,6 & 4 \\ a=4 & b=6 & 2 \\ \text { sum }=20 \end{array}$

$\begin{aligned} & \mathrm{n}(\mathrm{X})={ }^{20} \mathrm{C}_6={ }^{\mathrm{m}} \mathrm{C}_6 \\ & \mathrm{~m}=20 \end{aligned}$

given $|a-b| \geq 2$ so if

i.e. Total elements in X is ${ }^{20} \mathrm{C}_6$

Now for $\mathrm{n}(\mathrm{Y})$, range of R has exactly one element i.e. second elements must be constant in R and since R must have 6 element so it is not possible to satisfy both condition so $\mathrm{n}(\mathrm{Y})=0$.

$\begin{aligned} \text { for } \quad \mathrm{n}(\mathrm{z}) \quad & 1 \rightarrow 3,4,5,6 \\ & 2 \rightarrow 4,5,6 \\ & 3 \rightarrow 1,5,6 \\ & 4 \rightarrow 1,2,6 \\ & 5 \rightarrow 1,2,3 \\ & 6 \rightarrow 1,2,3,4 \end{aligned}$

no. of relation that are function will be

$\begin{aligned} & ={ }^4 \mathrm{C}_1 \times{ }^3 \mathrm{C}_1 \times{ }^3 \mathrm{C}_1 \times{ }^3 \mathrm{C}_1 \times{ }^3 \mathrm{C}_1 \times{ }^4 \mathrm{C}_1 \\ & =(4 \times 3 \times 3)^2=\mathrm{k}^2 \\ & \text { i.e. } \mathrm{k}=36 \end{aligned}$

$\matrix{ x & y & z \cr 2 & 3 & 4 \cr {{{\overline S }_1}} & {{{\overline S }_2}} & {} \cr }$

C-i) When x does not contain S$_1$, but contains S$_2$

$\mathop {{}^7{C_1}}\limits_{for\,x} \times \mathop {{{7!} \over {3!4!}}}\limits_{for\,y,z} = 245$

C-ii) When x does not contain $\mathrm{S}_1, \mathrm{~S}_2$ and y does not contain $\mathrm{S}_2$

i.e. $\mathop {{}^7{C_2}}\limits_{for\,x} \times \mathop {{{6!} \over {3!3!}}}\limits_{for\,y,z} = 420$

so total No. of ways 665

Let the 4-digit ATM pin code be represented by the digits $abxy$, where all digits are different, the greatest digit is 7, and the sum of the first two digits is equal to the sum of the last two digits: $a + b = x + y$.

Since the greatest digit is 7, the possible digits for the pin code are $0, 1, 2, 3, 4, 5, 6,$ and $7$.

1. Let's denote the sum of the first two digits and the sum of the last two digits as $\lambda$. Since the largest digit is 7, we can analyze different possible values for $\lambda$.

We will analyze different possible values for $\lambda$ and the corresponding possible pairs of digits:

- $\lambda = 7$: The pairs of digits that have a sum of 7 are $(0,7),(7,0),(1,6),(6,1),(2,5),(5,2),(3,4),(4,3)$. Since 7 must be present, we consider the pairs with 7: $(0,7),(7,0)$.

When the pair $(0,7)$ is chosen, the other two digits can be any one of the remaining pairs: $(1,6),(6,1),(2,5),(5,2),(3,4),(4,3)$. For each of these pairs, there are 2 possible pin codes: $07xy$ and $07yx$, where $x$ and $y$ are the digits from the chosen pair. In total, there are 12 possible pin codes for this case.

When the pair $(7,0)$ is chosen, the other two digits can also be any one of the remaining pairs: $(1,6),(6,1),(2,5),(5,2),(3,4),(4,3)$. For each of these pairs, there are 2 possible pin codes: $70xy$ and $70yx$, where $x$ and $y$ are the digits from the chosen pair. In total, there are 12 possible pin codes for this case.


So, there are 24 possible pin codes for $\lambda = 7$.



- $\lambda = 8$: The pairs of digits that have a sum of 8 are $(1,7),(7,1),(2,6),(6,2),(3,5),(5,3)$. Since 7 must be present, we consider the pairs with 7: $(1,7),(7,1)$.

When the pair $(1,7)$ is chosen, the other two digits can be any one of the remaining pairs: $(2,6),(6,2),(3,5),(5,3)$. For each of these pairs, there are 2 possible pin codes: $17xy$ and $17yx$, where $x$ and $y$ are the digits from the chosen pair. In total, there are 8 possible pin codes for this case.

When the pair $(7,1)$ is chosen, the other two digits can also be any one of the remaining pairs: $(2,6),(6,2),(3,5),(5,3)$. For each of these pairs, there are 2 possible pin codes: $71xy$ and $71yx$, where $x$ and $y$ are the digits from the chosen pair. In total, there are 8 possible pin codes for this case.

So, there are 16 possible pin codes for $\lambda = 8$.



- $\lambda = 9$: The pairs of digits that have a sum of 9 are $(2,7),(7,2),(3,6),(6,3),(4,5),(5,4)$. Since 7 must be present, we consider the pairs with 7: $(2,7),(7,2)$.

When the pair $(2,7)$ is chosen, the other two digits can be any one of the remaining pairs: $(3,6),(6,3),(4,5),(5,4)$. For each of these pairs, there are 2 possible pin codes: $27xy$ and $27yx$, where $x$ and $y$ are the digits from the chosen pair. In total, there are 8 possible pin codes for this case.

When the pair $(7,2)$ is chosen, the other two digits can also be any one of the remaining pairs: $(3,6),(6,3),(4,5),(5,4)$. For each of these pairs, there are 2 possible pin codes: $72xy$ and $72yx$, where $x$ and $y$ are the digits from the chosen pair. In total, there are 8 possible pin codes for this case.

So, there are 16 possible pin codes for $\lambda = 9$.



- $\lambda = 10$: The pairs of digits that have a sum of 10 are $(3,7),(7,3),(4,6),(6,4)$. Since 7 must be present, we consider the pairs with 7: $(3,7),(7,3)$.

When the pair $(3,7)$ is chosen, the other two digits can be any one of the remaining pairs: $(4,6),(6,4)$. For each of these pairs, there are 2 possible pin codes: $37xy$ and $37yx$, where $x$ and $y$ are the digits from the chosen pair. In total, there are 4 possible pin codes for this case.

When the pair $(7,3)$ is chosen, the other two digits can also be any one of the remaining pairs: $(4,6),(6,4)$. For each of these pairs, there are 2 possible pin codes: $73xy$ and $73yx$, where $x$ and $y$ are the digits from the chosen pair. In total, there are 4 possible pin codes for this case.

So, there are 8 possible pin codes for $\lambda = 10$.



- $\lambda = 11$: The pairs of digits that have a sum of 11 are $(4,7),(7,4),(5,6),(6,5)$. Since 7 must be present, we consider the pairs with 7: $(4,7),(7,4)$.

When the pair $(4,7)$ is chosen, the other two digits can be any one of the remaining pairs: $(5,6),(6,5)$. For each of these pairs, there are 2 possible pin codes: $47xy$ and $47yx$, where $x$ and $y$ are the digits from the chosen pair. In total, there are 4 possible pin codes for this case.

When the pair $(7,4)$ is chosen, the other two digits can also be any one of the remaining pairs: $(5,6),(6,5)$. For each of these pairs, there are 2 possible pin codes: $74xy$ and $74yx$, where $x$ and $y$ are the digits from the chosen pair. In total, there are 4 possible pin codes for this case.

So, there are 8 possible pin codes for $\lambda = 11$.



- $\lambda = 12$: The only pair of digits that have a sum of 12 is $(5,7)$. Since 7 must be present, this case is not possible.

- $\lambda = 13$: The only pair of digits that have a sum of 13 is $(6,7)$. Since 7 must be present, this case is not possible.

- $\lambda = 14$: The only pair of digits that have a sum of 14 is $(7,7)$. Since all the digits must be different, this case is not possible.

Adding up all the possible pin codes for each value of $\lambda$, we get:

$16 + 24 + 24 + 16 + 8 = 72$

Therefore, the maximum number of trials necessary to obtain the correct code is 72.

A number is divisible by 6 if it is divisible by both 2 and 3. A number is divisible by 2 if its last digit is even, which means it must be either 2 or 4 from the given digits. A number is divisible by 3 if the sum of its digits is divisible by 3.

$ \begin{aligned} & (2,1,3),(2,3,4),(2,5,5),(2,2,5),(2,2,2) \\\\ & (4,1,1),(4,4,1),(4,4,4),(4,3,5) \\\\ & 2,1,3 \Rightarrow 312,132 \\\\ & 2,3,4 \Rightarrow 342,432,234,324 \\\\ & 2,5,5 \Rightarrow 552 \\\\ & 2,2,5 \Rightarrow 252,522 \\\\ & 2,2,2 \Rightarrow 222 \\\\ & 4,1,1 \Rightarrow 114 \\\\ & 4,4,1 \Rightarrow 414,144 \\\\ & 4,4,4 \Rightarrow 444 \\\\ & 4,3,5 \Rightarrow 354,534 \end{aligned} $

Total 16 numbers.

$ x_1+x_2+x_3+\ldots x_7=12 $. This equation represents the number of ways to distribute 12 identical items (the sum of the digits) into 7 distinct boxes (the seven digits of the number), where each box can contain one of the numbers 1, 2, 3, or 4.

Number of solutions

$ \begin{aligned} & =\text { Coefficient of } x^{12} \text { in }\left(x^1+x^2+x^3+x^4\right)^7 \\\\ & =\text { Coefficient of } x^5 \text { in }\left(1+x+x^2+x^3\right)^7 \\\\ & =\text { Coefficient of } x^5 \text { in }\left(1-x^4\right)^7(1-x)^{-7} \\\\ & =\text { Coefficient of } x^5 \text { in }\left(1-7 x^4\right)(1-x)^{-7} \\\\ & =\text { Coefficient of } x^5 \text { in }\left(1-7 x^4\right) \sum_{r=0}^{\infty}{ }^{7+r-1} C_r \cdot x^r \\\\ & ={ }^{11} C_5-7 \times{ }^7 C_1 \\\\ & =462-49=413 \end{aligned} $

The problem involves forming nine-digit numbers from three digits a, b, c which are in Arithmetic Progression (AP), used three times each, such that at least once, three consecutive digits are in AP.

We have the two possible sequences for the AP :

1. a, b, c
2. c, b, a

This shows the flexibility in ordering the three digits that are in AP in our nine-digit number.

The next step is to choose the location of this sequence of three numbers within our nine-digit number.

Since there are nine places in our number and our sequence takes up three places, we have seven different starting points for our sequence : it can start at the first place, the second place, and so on, up to the seventh place.

Therefore, the number of ways to select 3 consecutive places out of the 9 places for the AP sequence is 7.

However, we also have to account for the fact that our sequence can be in one of two orders (a, b, c or c, b, a). So, we multiply the number of starting points by 2 to get $^7C_1 \times 2 = 14$ ways to arrange the sequence within our nine-digit number.

The remaining 6 digits (two 'a', two 'b', two 'c") can be arranged in $\frac{6!}{2!2!2!}$ ways.

Therefore, the total number of such nine-digit numbers is $^7C_1 \times 2 \times \frac{6!}{2!2!2!} = 1260$.

This problem can be solved using the concept of derangements, which is a permutation of objects where no object appears in its original position. In this case, we have 5 students who should not sit in their allotted seats.

The formula for calculating the number of derangements (also known as subfactorials) is :

D(n) $ =n !\left[\frac{1}{0!}-\frac{1}{1 !}+\frac{1}{2 !}-\frac{1}{3 !}+\frac{1}{4 !}-\ldots \ldots . .+(-1)^n \frac{1}{n !}\right] $

Where n is the number of students, in this case, 5.

Using the formula, let's calculate the derangements for 5 students :

D(5) = $5! \left(\frac{1}{0!} - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} - \frac{1}{5!}\right)$

D(5) = $120 \left(1 - 1 + \frac{1}{2} - \frac{1}{6} + \frac{1}{24} - \frac{1}{120}\right)$

D(5) = $120 \left(0 + \frac{1}{2} - \frac{1}{6} + \frac{1}{24} - \frac{1}{120}\right)$

D(5) = $120 \left(\frac{1}{2} - \frac{1}{6} + \frac{1}{24} - \frac{1}{120}\right)$

D(5) = $120 \left(\frac{60}{120} - \frac{20}{120} + \frac{5}{120} - \frac{1}{120}\right)$

D(5) = $120 \left(\frac{44}{120}\right)$

D(5) = 44

So, there are 44 ways in which none of the students sits on the allotted seat.

We have, four digits are $2,1,2,3$.

Total numbers when 1 is at unit digit $=\frac{3 !}{2 !}=3$

Total number when 2 is at unit digit $=3 !=6$

Total numbers when 3 is at unit digit $=\frac{3 !}{2 !}=3$

Sum of digits at unit place $=3 \times 1+6 \times 2+3 \times 3=24$

$\therefore$ Required sum $=24 \times 1000+24 \times 100+24 \times 10+24 \times 1$

$=24 \times 1111=26664$

Given that digits are $1,2,3,4,5,6,7$

Total permutations $=7$!

Let $p=$ Number which containing string 153

$q=$ Number which containing string 2467

$ \begin{array}{ll} & \therefore n(p)=5! \times 1 \\\\ & \Rightarrow n(q)=4! \times 1 \\\\ & \Rightarrow n(p \cap q)=2! \end{array} $

$ \begin{aligned} & \therefore n(p \cup q)=n(p)+n(q)-n(p \cap q) \\\\ & = 5 !+4 !-2 !=120+24-2=142 \end{aligned} $

$\therefore n$ (neither string 143 nor string 2467)

$ =7 !-142=5040-142=4898 $

Let, $n$ be the total number of couples who participated in the tournament.

According to the question, $2 \times{ }^n C_2 \times{ }^{n-2} C_2=840$

$ \begin{aligned} & \Rightarrow{ }^n C_2 \times{ }^{n-2} C_2=420 \\\\ & \Rightarrow \frac{n !}{2 !(n-2) !} \times \frac{(n-2) !}{(n-4) ! 2 !}=420 \\\\ & \Rightarrow \frac{n(n-1)(n-2)(n-3)}{4}=420 \end{aligned} $

Put $n=8$ satisfied the equation.

So, $n=8$

Hence, total number of players who participated
in the tournament $=2 \times 8=16$

Given, word is UNIVERSE

Here, vowels are E, I, U and consonants are N, R, S, V

$\therefore$ Required number of 4-letters words, with or without meaning,

each consisting of 2 vowels and 2 consonants

$ \begin{aligned} & ={ }^3 C_2 \times{ }^4 C_2 \times 4 ! \\\\ & ={ }^3 C_1 \times{ }^4 C_2 \times 24 \\\\ & =3 \times 6 \times 24=432 \end{aligned} $

Total ways without any restrictions :

There are $3^{20}$ ways to distribute the oranges to the 3 children.

Number of ways one child receives no orange :

Choose 1 child out of the 3 to not receive any orange in ${ }^3 C_1 = 3$ ways. Distribute 20 oranges to the remaining 2 children in $2^{20}$ ways. However, we've included the scenarios where the 2 children each get all the oranges. So, we subtract the 2 ways where one of the two remaining children gets all the oranges. $ { }^3 C_1(2^{20} - 2) $

Number of ways two children receive no orange :

Choose 2 children out of the 3 to not receive any oranges in ${ }^3 C_2 = 3$ ways. The third child will receive all 20 oranges in $1^{20} = 1$ way. $ { }^3 C_2 \times 1^{20} = 3 $

Number of ways

$=$ Total $-($ One child receive no orange + two child receive no orange)

$ \begin{aligned} & =3^{20}-\left({ }^3 C_1\left(2^{20}-2\right)+{ }^3 C_2 1^{20}\right) \\\\ & =3483638676 \end{aligned} $

$\begin{aligned} & x+y+z=21 \\\\ & \because \quad x \geq 1, y \geq 3, y \geq 4 \\\\ & \therefore \quad x_1+y_1+z_1=13 \\\\ & \text { Number of solutions }={ }^{13+3-1} C_{3-1} \\\\ & ={ }^{15} C_2=\frac{15 \times 14}{2}=7 \times 15 \\\\ & =105\end{aligned}$

A number will be divisible by 6 iff the digit at the unit place of the number is divisible by 2 and sum of all digits of the number is divisible by 3 .

Units, place must be occupied by 4 and hence, at least one 4 must be there.

Possible combination of 4, 5, 9 are as follows :

4	5	9	Total number of Number
1	1	4	${{5!} \over {4!}} = 5$
1	4	1	${{5!} \over {4!}} = 5$
2	2	2	${{5!} \over {2!2!}} = 30$
3	0	3	${{5!} \over {2!3!}} = 10$
3	3	0	${{5!} \over {2!3!}} = 10$
4	1	1	${{5!} \over {3!}} = 20$
6	0	0	${{5!} \over {5!}} = 1$

Total = 5 + 5 + 30 + 10 + 10 + 20 + 1 = 81.

$A=$ Numbers divisible by 2

$B=$ Numbers divisible by 3

$C=$ Numbers divisible by 7

$n(A \cup B)=n(A)+n(B)-n(A \cap B)$

$=n(2)+n(3)-n(6)$

$n(A)=n(2)=100,102 \ldots ., 998=450$

$n(B)=n(3)=102,105, \ldots ., 999=30$

$n(A \cap B)=n(6)=102,108, \ldots ., 996=150$

$n(2$ or 3$)=450+300-150=600$

Now,

$n(\mathrm{~A} \cap C)=n(14)=112,126, \ldots ., 994=64$

$n(A \cap B \cap C)=n(42)=126,168, \ldots ., 966=21$

$n(B \cap C)=n(21)=105,126, \ldots \ldots, 987,=43$

$n(2$ or 3 not by 7$)=600-[64+43-21]$

$=514$

Vowels : A,A,A,I,I,O

Consonants : S,S,S,S,N,N,T

$ \therefore $ Total number of ways in which vowels come together

$=\frac{8 !}{4 ! 2 !} \times \frac{6 !}{3 ! 2 !}=50400$

$A=\left[\begin{array}{ll}a_{11} & a_{12} \\ a_{13} & a_{14}\end{array}\right]$

Such that $\Sigma a_{i i}=3,5,7$ or 11

Then for sum 3, the possible entries are $(0,0,0,3)$, $(0,0,1,2),(0,1,1,1)$.

Then total number of possible matrices $ =4+12+4 $ $=20$

For sum 5 the possible entries are $(0,0,1,4)$, $(0,0,2,3),(0,1,2,2),(0,1,1,3)$ and $(1,1,1,2)$.

$\therefore $ Total possible matrices $=12+12+12+12+4=52$

For sum 7 the possible entries are $(0,0,3,4)$, $(0,2,2,3),(0,1,2,4),(0,1,3,3),(1,2,2,2)$, $(1,1,2,3)$ and $(1,1,1,4)$.

$\therefore $ Total possible matrices $=80$

For sum 11 the possible entries are $(0,3,4,4)$, $(1,2,4,4),(2,3,3,3),(2,2,3,4)$.

$\therefore $ Total number of matrices $=52$

$\therefore $ Total required matrices $=20+52+80+52$ $ =204 $

$\frac{\frac{(2 n+1) !}{(n+2) !}}{\frac{(2 n-1) !}{(n-1) !}}=\frac{11}{21}$

$\frac{(2 n+1) 2 n}{(n+2)(n+1) n}=\frac{11}{21}$

$84 n+42=11\left(n^{2}+3 n+2\right)$

$11 n^{2}-51 n-20=0$

$(n-5)(11 n+4)=0$

$n=5, \frac{-4}{11}$ (Rejected $)$

$n^{2}+n+15=45$

Numbers which are divisible by 3 (4 digit) and less than or equal to 2800

$=\frac{2799-1002}{3}+1=600$

Numbers which are divisible by 11 (4 digit) and less than or equal to 2800

$=\frac{2794-1001}{11}+1=164$

Numbers which are divisible by 33 (4 digit) and less than or equal to 2800

$=\frac{2772-1023}{33}+1=54$

$\therefore$ Total numbers = $600+164-54=710$

$.......1 \to {{6!} \over {2!3!}} = 60$

$.......3 \to {{6!} \over {3!}} = 120$

$.......5 \to {{6!} \over {3!2!}} = 60$

Total = 240

We have to make 4 digit numbers using the digits, 1, 2, 3 and 5.

The unit digit of the 4 digit number will be 5.

Now, the sum (x + y + z) should be of the (3p + 1).

Therefore, the possible cases are

(x, y, z) = (1, 1, 5), (1, 1, 2), (2, 2, 3), (2, 3, 5), (3, 3, 1) and (5, 5, 3).

So, total arrangements are

For $(1,1,5) \rightarrow \frac{3 !}{2 !}=3 ;$

For $(1,1,2) \rightarrow \frac{3 !}{2 !}=3 ;$

For $(2,2,3) \rightarrow \frac{3 !}{2 !}=3 ;$

For $(2,3,5) \rightarrow 3 !=6 ;$

For $(3,3,1) \rightarrow \frac{3 !}{2 !}=3 ;$

For $(5,5,3) \rightarrow \frac{3 !}{2 !}=3 ;$

So, total number of arrangements $=3+3+3+6+3+3=$ 21