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

2

3

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

1

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

1

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]

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

1

2

3

4

(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]

0%

(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]

0%

(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]

0%

(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]

0%

(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]

0%

(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]

0%

(43)

$Elements of the type 3 k = 3$

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

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

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

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

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

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

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

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

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

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

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]

0%

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]

0%

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]

0%

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]

0%

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]

0%

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]

0%

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] .question-row { display: flex; align-items: flex-start; gap: 8px; } .q-no { font-size: 17px; white-space: nowrap; margin-left: -15px; } .q-text { font-size: 18px; flex: 1; } .q-no, .q-text p { line-height: 30px; }

Q 4 : ∑k=06C351-k is equal to [2023] .question-row { display: flex; align-items: flex-start; gap: 8px; } .q-no { font-size: 17px; white-space: nowrap; margin-left: -15px; } .q-text { font-size: 18px; flex: 1; } .q-no, .q-text p { line-height: 30px; }

0%

0%

0%

0%

0%

0%

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 4 :

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