Permutation & Combination Model Questions Set 1 Section-Wise Topic Notes With Detailed Explanation And Example Questions
MOST IMPORTANT quantitative aptitude - 1 EXERCISES
The following question based on Permutations & Combination topic of quantitative aptitude
(a) $2/{11}$
(b) $1/2$
(c) $1/6$
(d) $3/7$
e) None of these
The correct answers to the above question in:
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 = ${n(E)}/{n(S)}$ = $6/{66}$ = $1/{11}$
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Read more model questions set 1 Based Quantitative Aptitude Questions and Answers
Question : 1
In how many different ways can 4 boys and 3 girls be arranged in a row such that all boys stand together and all the girls stand together?
a) 288
b) 576
c) 24
d) 75
e) None of these
Answer »Answer: (a)
Total number of ways to stand boys and girls together
= 4! × 3! × 2! = 4 × 3 × 2 × 3 × 2 × 2 = 288
Question : 2
Directions :
Answer these questions on the basis of the information given below : From a group of 6 men and 4 women a Committee of 4 persons is to be formed
In how many different ways can it be done so that the committee has at least one woman?
a) 195
b) 225
c) 185
d) 210
e) None of these
Answer »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
Question : 3
Only one number is wrong in the following number series. Find out the wrong number.
4, 6, 18, 49, 201, 1011
a) 201
b) 49
c) 1011
d) 18
e) None of these
Answer »Answer: (d)
The series is
× 1 + 2, × 2 + 3, × 3 + 4, × 4 + 5, × 5 + 6
The wrong number is 18.
It should be 6 × 2 + 3 = 15
Question : 4
In how many different ways can the letters of the word TRUST be arranged?
a) 80
b) 120
c) 25
d) 240
e) None of these
Answer »Answer: (e)
Required no. of ways = ${5!}/{2!}$ = 60 is
Total no. of letters in the word is 5; T is repeated twice.
Question : 5
If three marbles are picked at random, what is the probability that at least one is blue?
a) $5/{12}$
b) ${37}/{44}$
c) $7/{44}$
d) $7/{12}$
e) None of these
Answer »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 = $(1- {{35}/{220}})$ = $(1- {{7}/{55}})$ ${37}/{44}$
Question : 6
There are 4 candidates for the post of a lecturer in Mathematics and one is to be selected by votes of 5 men. The number of ways in which the votes can be given is
a) 40
b) 36
c) 48
d) 24
e) None of these
Answer »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
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