Permutation & Combination Model Questions Set 1 Section-Wise Topic Notes With Detailed Explanation And Example Questions

Answer: (e)

Total possible result = n (S)

=$^{12}C_2$ = ${12 × 11}/{1 × 2}$ = 66

Total number of event = n(E) = $^4C_2$ = ${4 × 3}/{1 × 2}$ = 6

∴ Required probability = ${n(E)}/{n(S)}$ = $6/{66}$ = $1/{11}$

Answer: (a)

Total number of ways to stand boys and girls together

= 4! × 3! × 2! = 4 × 3 × 2 × 3 × 2 × 2 = 288

Answer: (a)

The committee of 4 persons is to be so formed that it has at least 1 woman.

The different ways that we can choose to form such a committee are:

(i) 1w. 3 m in $^4C_1 × ^6C_3$ = 4 × ${6 × 5 × 4}/{3 × 2 × 1}$ = 80

(ii) 2w. 2 m in $^4C_2 × ^6C_2$ = ${4 × 3}/{2 × 1}$ × ${6 × 5}/{2 × 1}$ = 90

(iii) 3w. 1 m in $^4C_3 × ^6C_1$ = 4 × 6 = 24

(iv) 4w in $^4C_4$ = 1

∴ Total no. of different ways in which a committee of 4 persons can be formed so that it has at least one woman. = 80 + 90 + 24 + 1 = 195

Answer: (b)

Total possible result = n(S) = $^{12}C_3$ = ${12 × 11 × 10}/{1 × 2 × 3}$ = 220

Total number of event = n(E)

Except blue marbles, selection of 3 marbles out of 7 marbles

= $^7C_3$ = ${7 × 6 × 5}/{1 × 2 × 3}$ = 35

∴ Required probability = $(1- {{35}/{220}})$ = $(1- {{7}/{55}})$ ${37}/{44}$

Answer: (c)

Let the four candidates gets the votes x, y, z and w such that

x + y + z + w = 51 ...(i)

Here x ≥ 0 , y ≥ 0, z ≥ 0, w ≥ 0

The number of solutions of the above equation in this case is same as the number of ways in which the votes can be given if at least no two candidates get equal number of votes.

(Note : The number of ways in which n identical things can be distributed into r different groups = $^{n + r – 1}{C_{r - 1}}$)

∴ Total number of solutions of eqn. (i)

= $^{5 + 4 – 1}C_{4 – 1}$ = $^8C_3$ = 56

But in 8 ways the two candidate gets equal votes which are shown below :

(2, 2, 1, 0), (2, 2, 0, 1), (0, 2, 2, 1), (1, 2, 2, 0), (0, 1, 2, 2), (1, 0, 2, 2), (2, 0, 1, 2), (2, 1, 0, 2)

Hence the required number of ways = 56 – 8 = 48

Answer: (c)

Total number of ways to form committee

= $^5C_2 × ^6C_0 × ^3C_3 + ^5C_1 × ^6C_1 × ^3C_3 + ^5C_0 × ^6C_2 × ^3C_3$ = 10 + 30 + 15 = 55