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) $5/{12}$
(b) ${37}/{44}$
(c) $7/{44}$
(d) $7/{12}$
e) None of these
The correct answers to the above question in:
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 = $(1- {{35}/{220}})$ = $(1- {{7}/{55}})$ ${37}/{44}$
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Read more model questions set 1 Based Quantitative Aptitude Questions and Answers
Question : 1
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 : 2
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 : 3
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 : 4
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
Question : 5
the committee should include all the 3 Readers?
a) 21
b) 180
c) 55
d) 90
e) None of these
Answer »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
Question : 6
In how many ways can 7 persons be seated at a round table if 2 particular persons must not sit next to each other ?
a) 480
b) 240
c) 720
d) 5040
e) None of these
Answer »Answer: (a)
Total no. of unrestricted arrangements = (7 – 1) ! = 6 !
When two particular person always sit together, the total no. of arrangements = 6! – 2 × 5!
Required no. of arrangements = 6! – 2 × 5!
= 5! (6 – 2) = 5 × 4 × 3 × 2 × 4 = 480.
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