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
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
The correct answers to the above question in:
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
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Read more model questions set 1 Based Quantitative Aptitude Questions and Answers
Question : 1
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 : 2
What should come in place of the question mark (?) in the following number serie?
1, 3, 24, 360, 8640, 302400, ?
a) 154152000
b) 15425100
c) 14525100
d) 14515200
e) None of these
Answer »Answer: (d)
The series is as follows
×3, ×8, × 15, ×24, ×35, ×48
Hence, ? = 302400 × 48
= 14515200
Question : 3
What should come in place of the question mark (?) in the following number serie?
12, 14, 32, 102, 416, 2090, ?
a) 12552
b) 17552
c) 15522
d) 13525
e) None of these
Answer »Answer: (a)
The series is as follows
×1 + 2, ×2 + 4, ×3 + 6, ×4 + 8, × 5 + 10, × 6 + 12...
Hence, ? = 2090 × 6 + 12 = 12552
Question : 4
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 : 5
If two marbles are drawn at random, what is the probability that both are red?
a) $2/{11}$
b) $1/2$
c) $1/6$
d) $3/7$
e) None of these
Answer »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 = ${n(E)}/{n(S)}$ = $6/{66}$ = $1/{11}$
Question : 6
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.
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