Permutation & Combination Model Questions Set 1 Practice Questions Answers Test with Solutions & More Shortcuts

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: (e)

Required no. of ways = ${5!}/{2!}$ = 60 is

Total no. of letters in the word is 5; T is repeated twice.

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