Permutations and Combinations jee mains questions | JEE Main Maths Permutations and Combinations Previous Year Question

Q 1 :

If ${}^{2 n}{C_{3} : {}^{n}{C_{3} = 10 : 1,}}$ then the ratio $(n^{2} + 3 n) : (n^{2} - 3 n + 4)$ is [2023]

27 : 11
65 : 37
2 : 1
35 : 16

1

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4

(3)

We have, ${}^{2 n}{C_{3}} :^{n} C_{3} = 10 : 1$

$\Rightarrow \frac{(\frac{2 n!}{(2 n - 3)! 3!})}{\frac{n!}{3! (n - 3)!}} = \frac{10}{1}$

$\Rightarrow \frac{2 n (2 n - 1) (2 n - 2)}{6} \times \frac{6}{n (n - 1) (n - 2)} = \frac{10}{1}$

$\Rightarrow \frac{4 (2 n - 1)}{(n - 2)} = 10 \Rightarrow 8 n - 4 = 10 n - 20$

$\Rightarrow 16 = 2 n \Rightarrow n = 8$ $∴$ $(n^{2} + 3 n) : (n^{2} - 3 n + 4) = (64 + 24) : (64 - 24 + 4)$

$= 88 : 44 = 2 : 1$

Q 2 :

Let the number of elements in sets A and B be five and two respectively. Then the number of subsets of A x B each having at least 3 and at most 6 elements is [2023]

752
792
772
782

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2

3

4

(2)

$Given, n (A) = 5, n (B) = 2$

$∴$ $n (A \times B) = 10$

$∴$ $Required number of subsets =^{10} C_{3} +^{10} C_{4} +^{10} C_{5} +^{10} C_{6}$

$= 120 + 210 + 252 + 210$ $= 792$

Q 3 :

The number of triplets (x, y, z), where x, y, z are distinct non negative integers satisfying x + y + z = 15, is [2023]

80
114
92
136

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2

3

4

(2)

$x + y + z = 15$

$Total number of solutions =^{15 + 3 - 1} C_{3 - 1} = 136 ...(1)$

$Let x = y \neq z; 2 x + z = 15 \Rightarrow z = 15 - 2 x$

$\Rightarrow z \in {0, 1, 2, \dots, 7} - {5}$ $∴ There are 7 solutions.$

$∴ There are 21 solutions in which exactly two of x, y, z are equal. ...(2)$

$There is one solution in which x = y = z . ...(3)$

$Required answer = 136 - 21 - 1 = 114$

Q 4 :

$\sum_{k = 0}^{6} {}^{51 - k}{C_{3}}$ is equal to [2023]

${}^{51}{C_{3} -} {}^{45}{C_{3}}$
${}^{52}{C_{4} -} {}^{45}{C_{4}}$
${}^{52}{C_{3} -} {}^{45}{C_{3}}$
${}^{51}{C_{4} -} {}^{45}{C_{4}}$

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(2)

$\sum_{k = 0}^{6} {}^{51 - k}C_{3}$ $= {}^{51}C_{3} + {}^{50}C_{3} + {}^{49}C_{3} + {}^{48}C_{3} + {}^{47}C_{3} + {}^{46}C_{3} + {}^{45}C_{3}$

$As we have, {}^{n}C_{r} + {}^{n}C_{r - 1} = {}^{n + 1}C_{r} \Rightarrow {}^{n}C_{r - 1} = {}^{n + 1}C_{r} - {}^{n}C_{r}$

$So, {}^{51}C_{3} = {}^{52}C_{4} - {}^{51}C_{4}, {}^{50}C_{3} = {}^{51}C_{4} - {}^{50}C_{4}$ and ${}^{45}C_{3} = {}^{46}C_{4} - {}^{45}C_{4}$

$\Rightarrow \sum_{k = 0}^{6} {}^{51 - k}C_{3}$ $= {}^{52}C_{4} - {}^{51}C_{4} + {}^{51}C_{4} - {}^{50}C_{4} + \dots + {}^{47}C_{4} - {}^{46}C_{4} + {}^{46}C_{4} - {}^{45}C_{4}$

$= {}^{52}C_{4} - {}^{45}C_{4}$

Q 5 :

The number of ways of giving 20 distinct oranges to 3 children such that each child gets at least one orange is __________ . [2023]

(3483638676)

Required number of ways

$= {}^{3}{C_{0}} 3^{20} - {}^{3}{C_{1}} 2^{20} + {}^{3}{C_{2}} 1^{20}$

$= 3^{20} - 3 \times 2^{20} + 3$

$= 3 [3^{19} - 2^{20} + 1]$

$= 3483638676$

Q 6 :

The number of 4-letter words, with or without meaning, each consisting of 2 vowels and 2 consonants, which can be formed from the letters of the word UNIVERSE without repetition is ___________ . [2023]

(432)

$Vowels are U, I, E, E and consonants N, V, R, S$

$So, there are 4 vowels and 4 consonants.$

$When both vowels are different$

$∴$ $Number of ways = {}^{3}{C_{2}} \times {}^{4}{C_{2}} \times 4! = 432$

Q 7 :

Some couples participated in a mixed doubles badminton tournament. If the number of matches played, so that no couple played in a match, is 840, then the total numbers of persons, who participated in the tournament, is ___________ . [2023]

(16)

$Let the total number of couples = n .$

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

$\Rightarrow \frac{n!}{2! (n - 2)!} \times \frac{(n - 2)!}{2! (n - 4)!} \times 2 = 840$

$\Rightarrow \frac{n (n - 1) (n - 2) (n - 3)}{2} = 840 \Rightarrow n = 8$

$∴ Total number of players = 2 \times n = 2 \times 8 = 16$

Q 8 :

Number of integral solution to the equation x + y + z = 21, where $x \geq 1, y \geq 3, z \geq 4$ , is equal to __________ . [2023]

(105)

$Given, x + y + z = 21, x \geq 1, y \geq 3, z \geq 4$

$We have, x \geq 1, y \geq 3, z \geq 4$

$Let x - 1 = X,$ i.e., $x = 1 + X; y - 3 = Y,$ i.e., $y = Y + 3; z - 4 = Z,$ i.e., $z = Z + 4$ $∵$ $x + y + z = 21$

$\Rightarrow 1 + X + Y + 3 + Z + 4 = 21$

$∴$ $X + Y + Z = 21 - 8 = 13$

$∴$ $Number of non-negative integral solutions$

$= {}^{13 + 3 - 1}{C_{3 - 1}} = {}^{15}{C_{2}} = \frac{15 \times 14}{2} = 105$

Q 9 :

A boy needs to select five courses from 12 available courses, out of which 5 courses are language courses. If he can choose at most two language courses, then the number of ways he can choose five courses is _______ . [2023]

(546)

$5 language courses$

$7 non-language courses$

$He can select:$

$(i) 0 language courses and 5 non-language courses in {}^{5}{C_{0}} \times {}^{7}{C_{5}} = 21 ways$

$(ii) 1 language course and 4 non-language courses in {}^{5}{C_{1}} \times {}^{7}{C_{4}} = 175 ways$

$(iii) 2 language courses and 3 non-language courses in {}^{5}{C_{2}} \times {}^{7}{C_{3}} = 350 ways$

$Required number of ways = 21 + 175 + 350 = 546 ways$

Q 10 :

Let S = {1, 2, 3, 5, 7, 10, 11}. The number of non-empty subsets of S that have the sum of all elements a multiple of 3, is ________ . [2023]

(43)

$Elements of the type 3 k = 3$

$Elements of type 3 k + 1 = 1, 7, 10$

$Elements of type 3 k + 2 = 2, 5, 11$

$Subsets containing one element S_{1} = 1$

$Subsets containing two elements S_{2} = {}^{3}{C_{1}} \times {}^{3}{C_{1}} = 9$

$Subsets containing three elements S_{3} = {}^{3}{C_{1}} \times {}^{3}{C_{1}} + 1 + 1 = 11$

$Subsets containing four elements S_{4} = {}^{3}{C_{3}} + {}^{3}{C_{3}} + {}^{3}{C_{2}} \times {}^{3}{C_{2}} = 11$

$Subsets containing five elements S_{5} = {}^{3}{C_{2}} + {}^{3}{C_{2}} + {}^{3}{C_{2}} \times 1 = 9$

$Subsets containing six elements S_{6} = 1$

$Subsets containing seven elements S_{7} = 1$

$Required sum = 1 + 9 + 11 + 11 + 9 + 1 + 1 = 43$

Q 11 :

Suppose Anil's mother wants to give 5 whole fruits to Anil from a basket of 7 red apples, 5 white apples and 8 oranges. If in the selected 5 fruits, at least 2 oranges, at least one red apple and at least one white apple must be given, then the number of ways, Anil's mother can offer 5 fruits to Anil is _______ . [2023]

(6860)

$Let 7 red apples be denoted by (A), 5 white apples by (B), 8 oranges by (C) .$

$Now, 5 fruits are to be selected (Note: fruits taken different).$

$Possible selections: (2 C, 1 A, 2 B) or (2 C, 2 A, 1 B) or (3 C, 1 A, 1 B)$

$= {}^{8}{C_{2}} {}^{7}{C_{1}} {}^{5}{C_{2}} + {}^{8}{C_{2}} {}^{7}{C_{2}} {}^{5}{C_{1}} + {}^{8}{C_{3}} {}^{7}{C_{1}} {}^{5}{C_{1}}$

$= 28 \times 7 \times 10 + 28 \times 21 \times 5 + 56 \times 7 \times 5$

$= 1960 + 2940 + 1960 = 6860$

Q 12 :

If all the six digit numbers $x_{1} x_{2} x_{3} x_{4} x_{5} x_{6}$ with $0 < x_{1} < x_{2} < x_{3} < x_{4} < x_{5} < x_{6}$ are arranged in the increasing order, then the sum of the digits in the 72th number is ___________ . [2023]

(32)

2 4 5 6 7 8 → 72th word. So, 2 + 4 + 5 + 6 + 7 + 8 = 32

Q 13 :

Let the sets $S = {2, 4, 8, 16, \dots, 512}$ be partitioned into 3 sets $A, B, C$ with equal number of elements such that $A \cup B \cup C = S$ and $A \cap B = B \cap C = A \cap C = ϕ .$ The maximum number of such possible partitions of $S$ is equal to [2024]

1520
1640
1710
1680

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3

4

(4)

We have, $S = {2, 4, 8, 16, \dots, 512}$

$= {2^{1}, 2^{2}, 2^{3}, 2^{4}, \dots, 2^{9}}$ so, $| S | = 9$

$| A | = | B | = | C |$ and $A \cup B \cup C = S$

and $A \cap B = B \cap C = A \cap C = ϕ$

Since, $| S | = 9$ so $| A | = | B | = | C | = 3$

So, number of ways of making partition of $S = {}^{9}C_{3} \times {}^{6}{C_{3}} \times {}^{3}{C_{3}}$

$= \frac{9!}{6! 3!} \times \frac{6!}{3! \times 3!} \times 1$

$= \frac{9!}{3! \times 3! \times 3!} = 1680$

Q 14 :

Let $0 \leq r \leq n$ . If ${}^{n + 1}{C_{r + 1}} : {}^{n}{C_{r} : {}^{n - 1}{C_{r - 1}}} = 55 : 35 : 21,$ then $2 n + 5 r$ is equal to [2024]

55
50
60
62

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2

3

4

(2)

We have, ${}^{n + 1}{C_{r + 1} : {}^{n}{C_{r}}} = 55 : 35$

$\Rightarrow \frac{(n + 1)!}{(r + 1)! (n - r)!} \times \frac{(n - r)! r!}{n!} = \frac{55}{35}$

$\Rightarrow \frac{n + 1}{r + 1} = \frac{11}{7}$

$\Rightarrow 7 n = 11 r + 4$ ...(i)

Also, ${}^{n}{C_{r}} : {}^{n - 1}{C_{r - 1}} = 35 : 21$

$\Rightarrow \frac{n!}{r! (n - r)!} \times \frac{(r - 1)! (n - r)!}{(n - 1)!} = \frac{35}{21}$

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

$\Rightarrow 3 n = 5 r$ ...(ii)

Solving (i) and (ii), we get

$\Rightarrow 7 (\frac{5 r}{3}) = 11 r + 4$

$\Rightarrow 35 r - 33 r = 12$

$\Rightarrow r = 6$ and $n = 10$

So, $2 n + 5 r = 20 + 30 = 50$

Q 15 :

The number of ways five alphabets can be chosen from the alphabets of the word MATHEMATICS, where the chosen alphabets are not necessarily distinct, is equal to [2024]

175
181
179
177

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2

3

4

(3)

We have, MMAATTHEICS

Case-I : All distinct $= {}^{8}{C_{5}} = 56$

Case-II : 2 identical + 3 distinct $= {}^{3}{C_{1} \times {}^{7}{C_{3} = 105}}$

Case-III : 2 identical + 2 identical + 1 distinct $= {}^{3}{C_{2} \times {}^{6}{C_{1}} = 18}$

Hence, total number of ways = 56 + 105 + 18 = 179

Q 16 :

${}^{n - 1}{C_{r}} = (k^{2} - 8) {}^{n}{C_{r + 1}}$ if and only if [2024]

$2 \sqrt{2} < k \leq 3$
$2 \sqrt{3} < k \leq 3 \sqrt{2}$
$2 \sqrt{2} < k < 2 \sqrt{3}$
$2 \sqrt{3} < k < 3 \sqrt{3}$

1

2

3

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(1)

We have, ${}^{n - 1}{C_{r}} = (k^{2} - 8) {}^{n}{C_{r + 1}}$

$\Rightarrow \frac{(n - 1)!}{(n - (1 + r))! r!} = (k^{2} - 8) \frac{n!}{(n - (r + 1))! (r + 1)!}$

$\Rightarrow 1 = \frac{(k^{2} - 8) n}{r + 1} \Rightarrow k^{2} - 8 = \frac{r + 1}{n} \leq 1$ $[∵ n \geq r + 1]$

$\Rightarrow k^{2} - 8 \leq 1 \Rightarrow k^{2} - 9 \leq 0 \Rightarrow (k - 3) (k + 3) \leq 0$

$\Rightarrow - 3 \leq k \leq 3$ ...(i)

But $k^{2} > 8,$ as ${}^{n - 1}{C_{r}}$ can't be negative and 0.

$\Rightarrow k^{2} - 8 > 0 \Rightarrow (k - 2 \sqrt{2}) (k + 2 \sqrt{2}) > 0$

$\Rightarrow k < - 2 \sqrt{2} or k > 2 \sqrt{2}$ ...(ii)

From (i) and (ii), $2 \sqrt{2} < k \leq 3 or - 3 \leq k < - 2 \sqrt{2}$

Q 17 :

The number of ways in which 21 identical apples can be distributed among three children such that each child gets at least 2 apples, is [2024]

136
406
142
130

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2

3

4

(1)

We have, 21 identical apples.

Let $x_{1}, x_{2}$ and $x_{3}$ be the number of apples received by the three children.

$∴$ $x_{1} + x_{2} + x_{3} = 21$

Now, let $x_{1}^{'} = x_{1} - 2, x_{2}^{'} = x_{2} - 2$ and $x_{3}^{'} = x_{3} - 2$

$∴$ $x_{1}^{'} + x_{2}^{'} + x_{3}^{'} = 21 - 6 = 15$

$∴$ Required number of ways $= {}^{15 + 3 - 1}{C_{3 - 1}}$

$= {}^{17}{C_{2}} = 136$

Q 18 :

Let $a = 1 + \frac{{}^{2}{C_{2}}}{3!} + \frac{{}^{3}{C_{2}}}{4!} + \frac{{}^{4}{C_{2}}}{5!} + \dots,$ $b = 1 + \frac{{}^{1}{C_{0}} +^{1} C_{1}}{1!} + \frac{{}^{2}{C_{0}} +^{2} C_{1} +^{2} C_{2}}{2!} + \frac{{}^{3}{C_{0}} +^{3} C_{1} +^{3} C_{2} +^{3} C_{3}}{3!} + \dots$ Then $\frac{2 b}{a^{2}}$ is equal to ______ . [2024]

(8)

Given, $a = 1 + \frac{{}^{2}C_{2}}{3!} + \frac{{}^{3}C_{2}}{4!} + \frac{{}^{4}C_{2}}{5!} + \dots,$ $b = 1 + \frac{{}^{1}C_{0} + {}^{1}C_{1}}{1!} + \frac{{}^{2}C_{0} + {}^{2}C_{1} + {}^{2}C_{2}}{2!} + . . .,$

$\Rightarrow b = 1 + \frac{2}{1!} + \frac{2^{2}}{2!} + \frac{2^{3}}{3!} + \dots = e^{2}$ $[∵ e^{x} = 1 + \frac{x}{1!} + \frac{x^{2}}{2!} + \frac{x^{3}}{3!} + \dots]$

Now, $a = 1 + \sum_{r = 2}^{\infty} \frac{{}^{r}C_{2}}{(r + 1)!} = 1 + \sum_{r = 2}^{\infty} \frac{r (r - 1)}{2 (r + 1)!}$

$= 1 + \frac{1}{2} \sum_{r = 2}^{\infty} \frac{(r + 1) r - 2 r}{(r - 1)!} = 1 + \frac{1}{2} \sum_{r = 2}^{\infty} \frac{1}{(r - 1)!} - \frac{1}{2} \sum_{r = 2}^{\infty} \frac{2 r}{(r + 1)!}$

$= 1 + \frac{1}{2} (\frac{1}{1!} + \frac{1}{2!} + . . .) - \sum_{r = 2}^{\infty} \frac{(r + 1) - 1}{(r + 1)!}$

$= 1 + \frac{1}{2} (e - 1) - \sum_{r = 2}^{\infty} \frac{1}{r!} + \sum_{r = 2}^{\infty} \frac{1}{(r + 1)!}$

$= 1 + \frac{1}{2} (e - 1) - (e - \frac{1}{1!} - \frac{1}{0!}) + (e - \frac{1}{1!} - \frac{1}{0!} - \frac{1}{2!})$

$= 1 + \frac{e}{2} - \frac{1}{2} - e + 2 + e - 2 - \frac{1}{2} = \frac{e}{2} \Rightarrow \frac{2 b}{a^{2}} = \frac{2 e^{2}}{\frac{e^{2}}{4}} = 8$

Q 19 :

There are 4 men and 5 women in Group A, and 5 men and 4 women in Group B. If 4 persons are selected from each group, then the number of ways of selecting 4 men and 4 women is _______. [2024]

(5626)

Group A	Group B	Ways
$4 m$	$4 w$	${}^{4}{C_{4}} \cdot^{4} C_{4} = 1$
$3 m + 1 w$	$1 m + 3 w$	${}^{4}{C_{3}} \cdot^{5} C_{1} \cdot^{5} {C_{1}}^{4} C_{3} = 400$
$2 m + 2 w$	$2 m + 2 w$	${}^{4}{C_{2}} \cdot^{5} {C_{2}}^{5} {C_{2}}^{4} C_{2} = 3600$
$1 m + 3 w$	$3 m + w$	${}^{4}{C_{1}}^{5} {C_{3}}^{5} {C_{3}}^{4} C_{1} = 1600$
$4 w$	$4 m$	${}^{5}{C_{4}}^{5} C_{4} = 25$

Total number of ways = 1 + 400 + 3600 + 1600 + 25 = 5626

Q 20 :

The number of 3-digit numbers, formed using the digits 2, 3, 4, 5, and 7, when the repetition of digits is not allowed, and which are not divisible by 3, is equal to _______. [2024]

(36)

Possible triplets for which number is divisible by 3 are (2, 3, 4), (2, 3, 7), (3, 5, 4), (3, 5, 7)

These 4 triplets can be arrange in $4 \times 3! .$

Total number of 3-digit numbers made by using the digits 2, 3, 4, 5, and 7 $= {}^{5}{C_{3} \times 3!}$

Required number which are not divisible by 3 $= {}^{5}{C_{3} \times 3! - 4 \times 3! = 6 \times 3! = 36}$

Q 21 :

The lines $L_{1}, L_{2}, . . . L_{20}$ are distinct. For $n = 1, 2, 3, . . ., 10$ all the lines $L_{2 n - 1}$ are parallel to each other and all the lines $L_{2 n}$ pass through a given point P. The maximum number of points of intersection of pairs of lines from the set ${L_{1}, L_{2}, . . ., L_{20}}$ is equal to ______. [2024]

(101)

Let $L_{1}, L_{3}, . . ., L_{19}$ are parallel to each other and $L_{2}, L_{4}, L_{6}, \dots, L_{20}$ are passing through a point P.

$∴$ Point of intersection of pairs of lines from the set ${L_{1}, L_{2}, \dots, L_{20}} = {}^{20}{C_{2} - {}^{10}{C_{2}} - {}^{10}{C_{2}}} + 1 = 101$

Q 22 :

In an examination of Mathematics paper, there are 20 questions of equal marks and the question paper is divided into three sections: A, B, and C. A student is required to attempt total 15 questions taking at least 4 questions from each section. If section A has 8 questions, section B has 6 questions and section C has 6 questions, then the total number of ways a student can select 15 questions is ______. [2024]

(11376)

A B C Number of ways

5 6 4 ${}^{8}{C_{5}} \times {}^{6}{C_{6}} \times {}^{6}{C_{4}} = 56 \times 1 \times 15 = 840$

6 5 4 ${}^{8}{C_{6}} \times {}^{6}{C_{5}} \times {}^{6}{C_{4}} = 28 \times 6 \times 15 = 2520$

6 4 5 ${}^{8}{C_{6}} \times {}^{6}{C_{4}} \times {}^{6}{C_{5}} = 28 \times 15 \times 6 = 2520$

5 5 5 ${}^{8}{C_{5}} \times {}^{6}{C_{5}} \times {}^{6}{C_{5}} = 56 \times 6 \times 6 = 2016$

4 6 5 ${}^{8}{C_{4}} \times {}^{6}{C_{6}} \times {}^{6}{C_{5}} = 70 \times 1 \times 6 = 420$

4 5 6 ${}^{8}{C_{4}} \times {}^{6}{C_{5}} \times {}^{6}{C_{6}} = 70 \times 6 \times 1 = 420$

5 4 6 ${}^{8}{C_{5}} \times {}^{6}{C_{4}} \times {}^{6}{C_{6}} = 56 \times 15 \times 1 = 840$

7 4 4 ${}^{8}{C_{7}} \times {}^{6}{C_{4}} \times {}^{6}{C_{4}} = 8 \times 15 \times 15 = 1800$

$∴$ Required number of ways = 840 + 2520 + 2520 + 2016 + 420 + 420 + 840 = 11376

Q 23 :

The total number of words (with or without meaning) that can be formed out of the letters of the word 'DISTRIBUTION' taken four at a time, is equal to _______. [2024]

(3734)

The word 'DISTRIBUTION' has total 12 letters and 9 diferent letters.

We select four different letters.

Case l : Three same letters and 1 different letter ${}^{1}{C_{1}} \times {}^{8}{C_{1}} \times \frac{4!}{3!}$

Case 2 :Two pairs are same and two pairs are different ${}^{2}{C_{1}} \times {}^{8}{C_{2}} \times \frac{4!}{2!}$

Case 3 : Two pair of letters are same ${}^{2}{C_{2}} \cdot {}^{4}{C_{2}}$

Case 4 : All different letters are select ${}^{9}{C_{4}} \times 4!$

Total letters selected.

$= {}^{1}{C_{1}} \times {}^{8}{C_{1}} \times \frac{4!}{3!} + {}^{2}{C_{1}} \times {}^{8}{C_{2}} \cdot \frac{4!}{2!} + {}^{2}{C_{2}} \cdot {}^{4}{C_{2}} + {}^{9}{C_{4}} \cdot 4!$

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

$= 32 + 672 + 6 + 3024 = 3734$

Q 24 :

There are 5 points $P_{1}, P_{2}, P_{3}, P_{4}, P_{5}$ on the side AB, excluding A and B, of a triangle ABC. Similarly, there are 6 points $P_{6}, P_{7}, \dots, P_{11}$ on the side BC and 7 points $P_{12}, P_{13}, \dots, P_{18}$ on the side CA of the triangle. The number of triangles, that can be formed using the points $P_{1}, P_{2}, \dots, P_{18}$ as vertices, is: [2024]

796
751
771
776

1

2

3

4

(2)

Number of points on side AB = 5
Number of points on side BC = 6
Number of points on side AC = 7

Number of ways of selecting three points from side AB = ${}^{5}{C_{3}} $

Number of ways of selecting three points from side BC = ${}^{6}{C_{3}} $

Number of ways of selecting three points from side AC = ${}^{7}{C_{3}}$

Total number of triangles that can be formed using the points $P_{1}, P_{2}, \dots, P_{18} = {}^{18}{C_{3}} - ({}^{5}{C_{3} + {}^{6}{C_{3} + {}^{7}{C_{3}}}})$

$= {}^{18}{C_{3} - {}^{5}{C_{3}} - {}^{6}{C_{3}} - {}^{7}C_{3}} = 816 - 10 - 20 - 35 = 751$

Q 25 :

The number of triangles whose vertices are at the vertices of a regular octagon but none of whose sides is a side of the octagon is: [2024]

16
48
24
56

1

2

3

4

(1)

Total number of triangles that can be formed by using 8 vertices of the octagon = ${}^{8}{C_{3}}$

Number of triangles having exactly one side common with the octagon = $8 \times 4$

Since, if we choose AB as the common side, then other vertices of the triangle will be either of G, F, E or D as we have exactly one common side.

So, we have $8 \times 4$ i.e., $32$ such triangles.

Now, let us find the number of triangles having two common sides.

In the octagon, we have 8 ways of choosing two consecutive sides, i.e., (AB, BC), (BC, CD), (CD, DE), (DE, EF), (EF, FG), (FG, GH), (GH, HA), (HA, AB)

$∴$ Number of triangles having 2 sides common with the octagon = 8

$∴$ Required number of triangles = ${}^{8}{C_{3}} - 32 - 8$

$= 56 - 32 - 8 = 16$

Q 26 :

If for some $m, n;$ ${}^{6}{C_{m}} + 2 ({}^{6}{C_{m + 1}}) +^{6} C_{m + 2} >^{8} C_{3}$ and ${}^{n - 1}{P_{3}} :^{n} P_{4} = 1 : 8,$ then ${}^{n}{P_{m + 1}} +^{n + 1} C_{m}$ is equal to [2024]

372
384
380
376

1

2

3

4

(1)

Given, ${}^{6}{C_{m}} + 2 ({}^{6}{C_{m + 1}}) +^{6} C_{m + 2} >^{8} C_{3}$

$\Rightarrow^{6} C_{m} +^{6} C_{m + 1} +^{6} C_{m + 1} +^{6} C_{m + 2} >^{8} C_{3}$

$\Rightarrow^{7} C_{m + 1} +^{7} C_{m + 2} >^{8} C_{3} \Rightarrow^{8} C_{m + 2} >^{8} C_{3} {or}^{8} C_{6 - m} >^{8} C_{3}$

$\Rightarrow m + 2 > 3$ i.e., $m > 1$

or $6 - m > 3$ i.e., $m < 3$ i.e., $1 < m < 3 \Rightarrow m = 2$

$∴$ ${}^{n}{P_{m + 1}} +^{n + 1} C_{m} =^{8} P_{3} +^{9} C_{2} = 336 + 36 = 372$

Q 27 :

The number of ways, in which the letters A, B, C, D, E can be placed in the 8 boxes of the figure below so that no row remains empty and at most one letter can be placed in a box is :

[2025]

5880
5760
840
960

1

2

3

4

(2)

We have 5 letters A, B, C, D, E.

The following table shows the numbers of ways to fill 5 boxes so that no row remains empty.

Row-I	Row-II	Row-III	Number of ways
3	1	1	${}^{3}C_{3} \cdot {}^{3}C_{1} \cdot {}^{2}C_{1} = 6$
2	2	1	${}^{3}C_{2} \cdot {}^{3}C_{2} \cdot {}^{2}C_{1} = 18$
1	3	1	${}^{3}C_{1} \cdot {}^{3}C_{3} \cdot {}^{2}C_{1} = 6$
2	1	2	${}^{3}C_{2} \cdot {}^{3}C_{1} \cdot {}^{2}C_{2} = 9$
1	2	2	${}^{3}C_{1} \cdot {}^{3}C_{2} \cdot {}^{2}C_{2} = 9$

Number of ways to fill boxes = 6 + 18 + 6 + 9 + 9 = 48

Now, these 5 letters can be arranged in 5! ways

$∴$ Total number of ways $48 \times 5! = 5760$ .

Q 28 :

Line $L_{1}$ of slope 2 and line $L_{2}$ of slop $\frac{1}{2}$ intersect at the origin O. In the first quadrant, $P_{1}, P_{2}, . . ., P_{12}$ are 12 points on line $L_{1}$ and $Q_{1}, Q_{2}, . . ., Q_{9}$ are 9 points on line $L_{2}$ . Then the total number of triangles, that can be formed having vertices at three of the 22 points O, $P_{1}, P_{2}, . . ., P_{12}, Q_{1}, Q_{2}, . . ., Q_{9}$ is : [2025]

1080
1026
1134
1188

1

2

3

4

(3)

Total number of triangles are

$= {}^{9}C_{1} {}^{12}C_{2} + {}^{9}C_{2} {}^{12}C_{1} + {}^{1}C_{1} \cdot {}^{9}C_{1} {}^{12}C_{1}$

= 594 + 432 + 108 = 1134

Q 29 :

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 [2025]

145
155
165
135

1

2

3

4

(2)

To select 10 players including atleast 4 batsmen and 4 bowlers.

Captain and vice-captain already selected

$∴$ Total number of ways = ${}^{6}C_{5} \times {}^{5}C_{3} + {}^{6}C_{4} \times {}^{5}C_{4} + {}^{6}C_{3} \times {}^{5}C_{5}$

= $6 \times 10 + 15 \times 5 + 20 \times 1$

= 60 + 75 + 20 = 155

Q 30 :

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 [2025]

200
210
230
220

1

2

3

4

(2)

The number of ways to choose 3 points from 12 = ${}^{12}C_{3}$

= $\frac{12 \times 11 \times 10}{3 \times 2 \times 1}$ = 220

Number of ways to choose 3 points from 5 collinear points = ${}^{5}C_{3}$ = 10

Since these 5 points are collinear, so they can not form any triangles.

$∴$ Number of triangles that can be formed = 220 –10 = 210

Topic Question Set

Q 1 :

If ${}^{2 n}{C_{3} : {}^{n}{C_{3} = 10 : 1,}}$ then the ratio $(n^{2} + 3 n) : (n^{2} - 3 n + 4)$ is [2023]

Q 2 :

Let the number of elements in sets A and B be five and two respectively. Then the number of subsets of A x B each having at least 3 and at most 6 elements is [2023]

Q 3 :

The number of triplets (x, y, z), where x, y, z are distinct non negative integers satisfying x + y + z = 15, is [2023]

Q 4 :

$\sum_{k = 0}^{6} {}^{51 - k}{C_{3}}$ is equal to [2023]

Q 5 :

The number of ways of giving 20 distinct oranges to 3 children such that each child gets at least one orange is __________ . [2023]

Q 6 :

The number of 4-letter words, with or without meaning, each consisting of 2 vowels and 2 consonants, which can be formed from the letters of the word UNIVERSE without repetition is ___________ . [2023]

Q 7 :

Some couples participated in a mixed doubles badminton tournament. If the number of matches played, so that no couple played in a match, is 840, then the total numbers of persons, who participated in the tournament, is ___________ . [2023]

Q 8 :

Number of integral solution to the equation x + y + z = 21, where $x \geq 1, y \geq 3, z \geq 4$ , is equal to __________ . [2023]

Q 9 :

A boy needs to select five courses from 12 available courses, out of which 5 courses are language courses. If he can choose at most two language courses, then the number of ways he can choose five courses is _______ . [2023]

Q 10 :

Let S = {1, 2, 3, 5, 7, 10, 11}. The number of non-empty subsets of S that have the sum of all elements a multiple of 3, is ________ . [2023]

Q 12 :

If all the six digit numbers $x_{1} x_{2} x_{3} x_{4} x_{5} x_{6}$ with $0 < x_{1} < x_{2} < x_{3} < x_{4} < x_{5} < x_{6}$ are arranged in the increasing order, then the sum of the digits in the 72th number is ___________ . [2023]

Q 13 :

Let the sets $S = {2, 4, 8, 16, \dots, 512}$ be partitioned into 3 sets $A, B, C$ with equal number of elements such that $A \cup B \cup C = S$ and $A \cap B = B \cap C = A \cap C = ϕ .$ The maximum number of such possible partitions of $S$ is equal to [2024]

Q 14 :

Let $0 \leq r \leq n$ . If ${}^{n + 1}{C_{r + 1}} : {}^{n}{C_{r} : {}^{n - 1}{C_{r - 1}}} = 55 : 35 : 21,$ then $2 n + 5 r$ is equal to [2024]

Q 15 :

The number of ways five alphabets can be chosen from the alphabets of the word MATHEMATICS, where the chosen alphabets are not necessarily distinct, is equal to [2024]

Q 16 :

${}^{n - 1}{C_{r}} = (k^{2} - 8) {}^{n}{C_{r + 1}}$ if and only if [2024]

Q 17 :

The number of ways in which 21 identical apples can be distributed among three children such that each child gets at least 2 apples, is [2024]

Q 19 :

There are 4 men and 5 women in Group A, and 5 men and 4 women in Group B. If 4 persons are selected from each group, then the number of ways of selecting 4 men and 4 women is _______. [2024]

Q 20 :

The number of 3-digit numbers, formed using the digits 2, 3, 4, 5, and 7, when the repetition of digits is not allowed, and which are not divisible by 3, is equal to _______. [2024]

Q 23 :

The total number of words (with or without meaning) that can be formed out of the letters of the word 'DISTRIBUTION' taken four at a time, is equal to _______. [2024]

Q 25 :

The number of triangles whose vertices are at the vertices of a regular octagon but none of whose sides is a side of the octagon is: [2024]

Q 26 :

If for some $m, n;$ ${}^{6}{C_{m}} + 2 ({}^{6}{C_{m + 1}}) +^{6} C_{m + 2} >^{8} C_{3}$ and ${}^{n - 1}{P_{3}} :^{n} P_{4} = 1 : 8,$ then ${}^{n}{P_{m + 1}} +^{n + 1} C_{m}$ is equal to [2024]

Q 27 :

The number of ways, in which the letters A, B, C, D, E can be placed in the 8 boxes of the figure below so that no row remains empty and at most one letter can be placed in a box is :

[2025]

Q 30 :

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 [2025]

Want Us to Email you About Special Offers & Updates?

Topic Question Set

Q 1 : If C3:C3=10:1,n2n then the ratio (n2+3n):(n2-3n+4) is [2023]

Q 2 : Let the number of elements in sets A and B be five and two respectively. Then the number of subsets of A x B each having at least 3 and at most 6 elements is [2023]

Q 3 : The number of triplets (x, y, z), where x, y, z are distinct non negative integers satisfying x + y + z = 15, is [2023]

Q 4 : ∑k=06C351-k is equal to [2023]

Q 5 : The number of ways of giving 20 distinct oranges to 3 children such that each child gets at least one orange is __________ . [2023]

Q 6 : The number of 4-letter words, with or without meaning, each consisting of 2 vowels and 2 consonants, which can be formed from the letters of the word UNIVERSE without repetition is ___________ . [2023]

Q 7 : Some couples participated in a mixed doubles badminton tournament. If the number of matches played, so that no couple played in a match, is 840, then the total numbers of persons, who participated in the tournament, is ___________ . [2023]

Q 8 : Number of integral solution to the equation x + y + z = 21, where x≥1, y≥3, z≥4, is equal to __________ . [2023]

Q 9 : A boy needs to select five courses from 12 available courses, out of which 5 courses are language courses. If he can choose at most two language courses, then the number of ways he can choose five courses is _______ . [2023]

Q 10 : Let S = {1, 2, 3, 5, 7, 10, 11}. The number of non-empty subsets of S that have the sum of all elements a multiple of 3, is ________ . [2023]

Q 12 : If all the six digit numbers x1 x2 x3 x4 x5 x6 with 0<x1<x2<x3<x4<x5<x6 are arranged in the increasing order, then the sum of the digits in the 72th number is ___________ . [2023]

Q 13 : Let the sets S={2,4,8,16,…,512} be partitioned into 3 sets A,B,C with equal number of elements such that A∪B∪C=S and A∩B=B∩C=A∩C=ϕ. The maximum number of such possible partitions of S is equal to [2024]

Q 14 : Let 0≤r≤n. If Cr+1n+1:Cr:Cr-1n-1n=55:35:21, then 2n+5r is equal to [2024]

Q 15 : The number of ways five alphabets can be chosen from the alphabets of the word MATHEMATICS, where the chosen alphabets are not necessarily distinct, is equal to [2024]

Q 16 : Crn-1=(k2-8)Cr+1n if and only if [2024]

Q 17 : The number of ways in which 21 identical apples can be distributed among three children such that each child gets at least 2 apples, is [2024]

Q 18 : Let a=1+C223!+C234!+C245!+…,b=1+C01+1C11!+C02+2C1+2C22!+C03+3C1+3C2+3C33!+… Then 2ba2 is equal to ______ . [2024]

Q 19 : There are 4 men and 5 women in Group A, and 5 men and 4 women in Group B. If 4 persons are selected from each group, then the number of ways of selecting 4 men and 4 women is _______. [2024]

Q 20 : The number of 3-digit numbers, formed using the digits 2, 3, 4, 5, and 7, when the repetition of digits is not allowed, and which are not divisible by 3, is equal to _______. [2024]

Q 21 : The lines L1,L2,...L20 are distinct. For n=1,2,3,...,10 all the lines L2n-1 are parallel to each other and all the lines L2n pass through a given point P. The maximum number of points of intersection of pairs of lines from the set {L1,L2,...,L20} is equal to ______. [2024]

Q 23 : The total number of words (with or without meaning) that can be formed out of the letters of the word 'DISTRIBUTION' taken four at a time, is equal to _______. [2024]

Q 25 : The number of triangles whose vertices are at the vertices of a regular octagon but none of whose sides is a side of the octagon is: [2024]

Q 26 : If for some m,n; Cm6+2(Cm+16)+6Cm+2>8C3 and P3n-1:nP4=1:8, then Pm+1 n+ n+1Cm is equal to [2024]

Q 27 : The number of ways, in which the letters A, B, C, D, E can be placed in the 8 boxes of the figure below so that no row remains empty and at most one letter can be placed in a box is : [2025]

Q 30 : 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 [2025]

Want Us to Email you About Special Offers & Updates?

Q 1 :

If ${}^{2 n}{C_{3} : {}^{n}{C_{3} = 10 : 1,}}$ then the ratio $(n^{2} + 3 n) : (n^{2} - 3 n + 4)$ is [2023]

Q 2 :

Let the number of elements in sets A and B be five and two respectively. Then the number of subsets of A x B each having at least 3 and at most 6 elements is [2023]

Q 3 :

The number of triplets (x, y, z), where x, y, z are distinct non negative integers satisfying x + y + z = 15, is [2023]

Q 4 :

$\sum_{k = 0}^{6} {}^{51 - k}{C_{3}}$ is equal to [2023]

Q 5 :

The number of ways of giving 20 distinct oranges to 3 children such that each child gets at least one orange is __________ . [2023]

Q 6 :

The number of 4-letter words, with or without meaning, each consisting of 2 vowels and 2 consonants, which can be formed from the letters of the word UNIVERSE without repetition is ___________ . [2023]

Q 7 :

Some couples participated in a mixed doubles badminton tournament. If the number of matches played, so that no couple played in a match, is 840, then the total numbers of persons, who participated in the tournament, is ___________ . [2023]

Q 8 :

Number of integral solution to the equation x + y + z = 21, where $x \geq 1, y \geq 3, z \geq 4$ , is equal to __________ . [2023]

Q 9 :

A boy needs to select five courses from 12 available courses, out of which 5 courses are language courses. If he can choose at most two language courses, then the number of ways he can choose five courses is _______ . [2023]

Q 10 :

Let S = {1, 2, 3, 5, 7, 10, 11}. The number of non-empty subsets of S that have the sum of all elements a multiple of 3, is ________ . [2023]

Q 12 :

If all the six digit numbers $x_{1} x_{2} x_{3} x_{4} x_{5} x_{6}$ with $0 < x_{1} < x_{2} < x_{3} < x_{4} < x_{5} < x_{6}$ are arranged in the increasing order, then the sum of the digits in the 72th number is ___________ . [2023]

Q 13 :

Let the sets $S = {2, 4, 8, 16, \dots, 512}$ be partitioned into 3 sets $A, B, C$ with equal number of elements such that $A \cup B \cup C = S$ and $A \cap B = B \cap C = A \cap C = ϕ .$ The maximum number of such possible partitions of $S$ is equal to [2024]

Q 14 :

Let $0 \leq r \leq n$ . If ${}^{n + 1}{C_{r + 1}} : {}^{n}{C_{r} : {}^{n - 1}{C_{r - 1}}} = 55 : 35 : 21,$ then $2 n + 5 r$ is equal to [2024]

Q 15 :

The number of ways five alphabets can be chosen from the alphabets of the word MATHEMATICS, where the chosen alphabets are not necessarily distinct, is equal to [2024]

Q 16 :

${}^{n - 1}{C_{r}} = (k^{2} - 8) {}^{n}{C_{r + 1}}$ if and only if [2024]

Q 17 :

The number of ways in which 21 identical apples can be distributed among three children such that each child gets at least 2 apples, is [2024]

Q 19 :

There are 4 men and 5 women in Group A, and 5 men and 4 women in Group B. If 4 persons are selected from each group, then the number of ways of selecting 4 men and 4 women is _______. [2024]

Q 20 :

The number of 3-digit numbers, formed using the digits 2, 3, 4, 5, and 7, when the repetition of digits is not allowed, and which are not divisible by 3, is equal to _______. [2024]

Q 23 :

The total number of words (with or without meaning) that can be formed out of the letters of the word 'DISTRIBUTION' taken four at a time, is equal to _______. [2024]

Q 25 :

The number of triangles whose vertices are at the vertices of a regular octagon but none of whose sides is a side of the octagon is: [2024]

Q 26 :

If for some $m, n;$ ${}^{6}{C_{m}} + 2 ({}^{6}{C_{m + 1}}) +^{6} C_{m + 2} >^{8} C_{3}$ and ${}^{n - 1}{P_{3}} :^{n} P_{4} = 1 : 8,$ then ${}^{n}{P_{m + 1}} +^{n + 1} C_{m}$ is equal to [2024]

Q 27 :

The number of ways, in which the letters A, B, C, D, E can be placed in the 8 boxes of the figure below so that no row remains empty and at most one letter can be placed in a box is :

[2025]

Q 30 :

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 [2025]