Permutations and Combinations
Three persons enter in a lift at the ground floor. The lift will go up to 10th floor. The number of ways, in which the three persons can exit the lift at three different floors, if the lift does not stop at first, second and third floors, is equal to ________.
Explanation:
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 $
Explanation:
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 = $8+16+18=42$
Let S denote the set of 4-digit numbers $a b c d$ such that $a>b>c>d$ and P denote the set of 5 -digit numbers having product of its digits equal to 20 . Then $n(\mathrm{~S})+n(\mathrm{P})$ is equal to $\_\_\_\_$
Explanation:
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 .
The number of 4 -letter words, with or without meaning, which can be formed using the letters PQRPQRSTUVP, is $\_\_\_\_$ .
Explanation:
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} $
Let ABC be a triangle. Consider four points $\mathrm{p}_1, \mathrm{p}_2, \mathrm{p}_3, \mathrm{p}_4$ on the side AB , five points $p_5, p_6, p_7, p_8, p_9$ on the side $B C$, and four points $p_{10}, p_{11}, p_{12}, p_{13}$ on the side AC . None of these points is a vertex of the triangle ABC . Then the total number of pentagons, that can be formed by taking all the vertices from the points $p_1, p_2, \ldots, p_{13}$, is $\_\_\_\_$
Explanation:
$ { }^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 $
Let $S=\{(m, n): m, n \in\{1,2,3, \ldots . ., 50\}\}$. If the number of elements $(m, n)$ in $S$ such that $6^m+9^n$ is a multiple of 5 is $p$ and the number of elements ( $m, n$ ) in $S$ such that $m+n$ is a square of a prime number is q , then $\mathrm{p}+\mathrm{q}$ is equal to $\_\_\_\_$ .
Explanation:
$ \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
$\Rightarrow 1333$
Let $\mathrm{S}=\{1,2,3,4,5,6,7,8,9\}$. Let $x$ be the number of 9-digit numbers formed using the digits of the set S such that only one digit is repeated and it is repeated exactly twice. Let $y$ be the number of 9 -digit numbers formed using the digits of the set S such that only two digits are repeated and each of these is repeated exactly twice. Then,
$56 x=9 y$
$21 x=4 y$
$45 x=7 y$
$29 x=5 y$
The letters of the word "UDAYPUR" are written in all possible ways with or without meaning and these words are arranged as in a dictionary. The rank of the word "UDAYPUR" is
1579
1578
1580
1581
The largest value of $n$, for which $40^n$ divides $60!$, is
14
13
11
12
The number of ways, in which 16 oranges can be distributed to four children such that each child gets at least one orange, is
384
403
429
455
The largest $n \in \mathbb{N}$, for which $7^n$ divides $101!$, is :
18
15
19
16
The number of strictly increasing functions $f$ from the set $\{1,2,3,4,5,6\}$ to the set $\{1,2,3, \ldots ., 9\}$ such that $f(i) \neq i$ for $1 \leq i \leq 6$, is equal to :
21
28
27
22
Let $S = \{1, 2, 3, \ldots, 10\}$. Consider the set
$X = \{R : R \text{ is an equivalence relation on the set } S \text{ such that } R \text{ has exactly 42 elements}\}$.
Then the number of elements in $X$ is ____________.
Explanation:
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 $
The number of ways to distribute 10 identical red pens and 14 identical blue pens among four persons such that each person gets 6 pens, is ______________.
Explanation:
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.
Let m and $\mathrm{n},(\mathrm{m}<\mathrm{n})$, be two 2-digit numbers. Then the total numbers of pairs $(\mathrm{m}, \mathrm{n})$, such that $\operatorname{gcd}(m, n)=6$, is __________ .
Explanation:
$\begin{aligned} & m=6 a, n=6 b \\ & \text { So } \operatorname{gcd}(m, n)=6 \Rightarrow \operatorname{gcd}(a, b)=1 \\ & m=6 a \geq 10 \Rightarrow a \geq\left[\frac{10}{6}\right]=2 \\ & m=6 a \leq 99 \Rightarrow a \leq\left[\frac{99}{6}\right]=16 \end{aligned}$
So $a, b \in\{2,3, \ldots, 16\}$, and we count how many coprime pairs $(a, b)$ with $a< b, \operatorname{gcd}(a, b)=1$
$\begin{aligned} & a=2 \Rightarrow b=3,5,7,9,11,13,15 \Rightarrow 7 \\ & a=3 \Rightarrow \mathrm{~b}=4,5,7,8,10,11,13,14,16 \Rightarrow 9 \\ & a=4 \Rightarrow b=5,7,9,11,13,15 \Rightarrow 6 \\ & a=5 \Rightarrow \mathrm{~b}=6,7,8,9,11,12,13,14,16 \Rightarrow 9 \\ & a=6 \Rightarrow \mathrm{~b}=7,11,13 \Rightarrow 3 \\ & a=7 \Rightarrow b=8,9,10,11,12,13,15,16 \Rightarrow 8 \\ & a=8 \Rightarrow \mathrm{~b}=9,11,13,15 \Rightarrow 4 \\ & a=9 \Rightarrow \mathrm{~b}=10,11,13,14,16 \Rightarrow 5 \\ & a=10 \Rightarrow b=11,13 \Rightarrow 2 \\ & a=11 \Rightarrow b=12,13,14,15,16 \Rightarrow 5 \\ & a=12 \Rightarrow b=13,17 \times \rightarrow \text { only } 13 \text { is valid } \Rightarrow 1 \\ & a=13 \Rightarrow b=14,15,16 \Rightarrow 3 \\ & a=14 \Rightarrow b=15, \Rightarrow 1 \\ & a=15 \Rightarrow b=16 \Rightarrow 1 \end{aligned}$
$\begin{aligned} & \text { Total }=7+9+6+9+3+8+4+5+2+5+1 \\ & +3+1+1=64\end{aligned}$
All five letter words are made using all the letters A, B, C, D, E and arranged as in an English dictionary with serial numbers. Let the word at serial number $n$ be denoted by $\mathrm{W}_{\mathrm{n}}$. Let the probability $\mathrm{P}\left(\mathrm{W}_{\mathrm{n}}\right)$ of choosing the word $\mathrm{W}_{\mathrm{n}}$ satisfy $\mathrm{P}\left(\mathrm{W}_{\mathrm{n}}\right)=2 \mathrm{P}\left(\mathrm{W}_{\mathrm{n}-1}\right), \mathrm{n}>1$.
If $\mathrm{P}(\mathrm{CDBEA})=\frac{2^\alpha}{2^\beta-1}, \alpha, \beta \in \mathbb{N}$, then $\alpha+\beta$ is equal to :____________
Explanation:
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 $.
If the number of seven-digit numbers, such that the sum of their digits is even, is $m \cdot n \cdot 10^n ; m, n \in\{1,2,3, \ldots, 9\}$, then $m+n$ is equal to__________
Explanation:
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 $9 \cdot 5 \cdot 10^5=4.5 \cdot 10^6$
$m+n=9+5=14$
The number of 6-letter words, with or without meaning, that can be formed using the letters of the word MATHS such that any letter that appears in the word must appear at least twice, is _________.
Explanation:
(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 $=1405$
The number of natural numbers, between 212 and 999, such that the sum of their digits is 15, is _______.
Explanation:
$\begin{array}{|c|c|c|} \hline \mathrm{x} & \mathrm{y} & \mathrm{z} \\ \hline \end{array}$
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 $=6=7+8+9+10+9+8+7=64$
Number of functions $f:\{1,2, \ldots, 100\} \rightarrow\{0,1\}$, that assign 1 to exactly one of the positive integers less than or equal to 98 , is equal to ________.
Explanation:
392 Ans.
The number of 3 -digit numbers, that are divisible by 2 and 3 , but not divisible by 4 and 9 , is _________.
Explanation:
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 $4 \& 9$
$150-25=125$
The number of ways, 5 boys and 4 girls can sit in a row so that either all the boys sit together or no two boys sit together, is ________.
Explanation:
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 $=5!\times 5!+4!\times 5!=17280$
There are 12 points in a plane, no three of which are in the same straight line, except 5 points which are collinear. Then the total number of triangles that can be formed with the vertices at any three of these 12 points is
230
210
200
220
From a group of 7 batsmen and 6 bowlers, 10 players are to be chosen for a team, which should include atleast 4 batsmen and atleast 4 bowlers. One batsmen and one bowler who are captain and vice-captain respectively of the team should be included. Then the total number of ways such a selection can be made, is
The number of sequences of ten terms, whose terms are either 0 or 1 or 2 , that contain exactly five 1 s and exactly three 2 s , is equal to :
If all the words with or without meaning made using all the letters of the word "KANPUR" are arranged as in a dictionary, then the word at 440th position in this arrangement is :
PRNAKU
PRKAUN
PRKANU
PRNAUK
Let $ P $ be the set of seven digit numbers with sum of their digits equal to 11. If the numbers in $ P $ are formed by using the digits 1, 2 and 3 only, then the number of elements in the set $ P $ is :
164
158
161
173
Let ${ }^n C_{r-1}=28,{ }^n C_r=56$ and ${ }^n C_{r+1}=70$. Let $A(4 \operatorname{cost}, 4 \sin t), B(2 \sin t,-2 \cos t)$ and $C\left(3 r-n, r^2-n-1\right)$ be the vertices of a triangle $A B C$, where $t$ is a parameter. If $(3 x-1)^2+(3 y)^2$ $=\alpha$, is the locus of the centroid of triangle ABC , then $\alpha$ equals
The number of different 5 digit numbers greater than 50000 that can be formed using the digits 0 , $1,2,3,4,5,6,7$, such that the sum of their first and last digits should not be more than 8 , is
Group A consists of 7 boys and 3 girls, while group B consists of 6 boys and 5 girls. The number of ways, 4 boys and 4 girls can be invited for a picnic if 5 of them must be from group $A$ and the remaining 3 from group $B$, is equal to :
The number of words, which can be formed using all the letters of the word "DAUGHTER", so that all the vowels never come together, is :
In a group of 3 girls and 4 boys, there are two boys $B_1$ and $B_2$. The number of ways, in which these girls and boys can stand in a queue such that all the girls stand together, all the boys stand together, but $B_1$ and $B_2$ are not adjacent to each other, is :
From all the English alphabets, five letters are chosen and are arranged in alphabetical order. The total number of ways, in which the middle letter is ' M ', is :
Let the set of all relations $R$ on the set $\{a, b, c, d, e, f\}$, such that $R$ is reflexive and symmetric, and $R$ contains exactly $10$ elements, be denoted by $\mathcal{S}$.
Then the number of elements in $\mathcal{S}$ is ________________.
Explanation:
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} $
Let $S$ be the set of all seven-digit numbers that can be formed using the digits $0, 1$ and $2$. For example, $2210222$ is in $S$, but $0210222$ is NOT in $S$.
Then the number of elements $x$ in $S$ such that at least one of the digits $0$ and $1$ appears exactly twice in $x$, is equal to ____________.
All the letters of the word MOTHER are arranged in all possible ways and the resulting words (may or may not have meaning) are arranged as in the dictionary. The number of words that appear after the word MOTHER is
309
310
410
411
The number of positive integral solution of $\frac{1}{x}+\frac{1}{y}=\frac{1}{2025}$ is
105
45
135
25
The number of positive integral solutions of $x y z=60$ is
${ }^{59} \mathrm{C}_2$
${ }^4 \mathrm{C}_2 \times{ }^3 \mathrm{C}_2 \times{ }^3 \mathrm{C}_2$
${ }^4 \mathrm{C}_3$
${ }^3 \mathrm{C}_1 \times{ }^4 \mathrm{C}_0 \times{ }^4 \mathrm{C}_4$
5 boys and 5 girls have to sit around a table. The number of ways in which all of them can sit so that no two boys and no two girls are together is
14400
2880
576
625
$⇒ Number of ways to arrange 5 boys and 5 girls in a circle, so that no two boys and no two girls are together
All possible words (with or without meaning) the contain the word 'GENTLE' are formed using all the letters of the word 'INTELLIGENCE'. Then, the number of words in which the word 'GENTLE' appears among the first nine positions only is
1440
5040
2520
720
$ { }^{20} P_5-{ }^{19} P_5= $
${ }^{19} P_4$
$4\left({ }^{19} P_4\right)$
$5!(646)$
$6!(646)$
If all the letters of the word ACADEMICIAN are permuted in all possible ways, then the number of permutations in which no two $A^{\prime} s$ are together and all the consonants are together is
7200
14400
3600
1800
The number of all possible three letter words that can be formed by choosing three letters from the letters of the word FEBRUARY so that a vowel always occupies the middle place is
90
93
126
129
The number of ways in which 6 boys and 4 girls can be arranged in a row such that between any two girls there must be exactly 2 boys is
$6!5!$
(72)6!
$(144) 5$ !
$4!7!$
There are 15 stations on a train route and the train has to be stopped at exactly 5 stations among these 15 stations. If it stops at atleast two consecutive stations, then the number of ways in which the train can be stopped is
${ }^{11} \mathrm{C}_5$
${ }^{15} \mathrm{C}_5$
${ }^{15} \mathrm{C}_5-{ }^{11} \mathrm{C}_5$
${ }^{15} \mathrm{C}_{10}-{ }^9 \mathrm{C}_5$
Number of all possible ways of distributing eight identical apples among three persons is
45
42
39
36
Number of all possible words (with or without meaning) that can be formed using all the letters of the word CABINET in which neither the word CAB nor the word NET appear is
5040
4806
4800
5034
The number of non-negative integral solutions of the equation $x+y+z+t=10$ when $x \geq 2, z \geq 5$ is
80
20
50
10