This set of Aptitude Questions and Answers (MCQs) focuses on “Permutations and Combinations – Set 3”.
1. A set consists of 12 different words. In how many ways can we choose a subset of not more than 4 words?
a) 794
b) 840
c) 12C4
d) 12! * 4!
View Answer
Explanation: Given, the set has 12 different words.
We have to choose a subset of 4 words from the given.
12C0 + 12C1 + 12C2 + 12C3 + 12C4 = 1 + 12 + 66 + 220 + 495 = 794
2. How many numbers can be formed with even numbers 2, 4, 6, 8 without repetition?
a) 24
b) 64
c) 24
d) 48
View Answer
Explanation: Number of one digit numbers we can form = 4.
Number of two digit numbers we can form = 4 * 3 = 12.
Number of three digit numbers we can form = 4 * 3 * 2 = 24.
Number of four digit numbers we can form = 4 * 3 * 2 * 1 = 24.
Total number of numbers we can form = 4 + 12 + 24 + 24 = 64.
3. Sony’s class teacher informed her to select 6 out of 10 subjects, and also she told her that two of the subjects are compulsory to choose and the rest of them are of your interest. In how many ways Sony can choose her subjects?
a) 6!
b) 35
c) 70
d) 10!
View Answer
Explanation: Given that, there are 10 subjects in which Sony should select 6.
It is given that 2 of them are compulsorily to choose.
If two subjects are compulsory to choose then there will be 8 subjects left.
In the 8 subjects Sony should choose the other 4 subjects.
Sony can select the subjects in 8C4 = 70 ways.
4. In how many ways can the word ‘REMOVE’ be arranged, when repetition is not allowed?
a) 6! * 2!
b) 66 / 2!
c) 6!
d) 6! / 2!
View Answer
Explanation: The word ‘REMOVE’ has 6 letters.
It is given that repetition is not allowed.
In this the letter ‘E’ is repeated.
Let’s assume x = number of letters in the word.
y = (number of times the letter is repeated)!.
Number of ways the word can be arranged = x! / y! = 6! / 2!.
5. There are 15 employees working on a project. The project is successfully approved by the company and the company decided to give the best employee award. In order to give it they have to make comparisons among the 15 members. Find the number of comparisons needed?
a) 15! / 2!
b) 15!
c) 100
d) 105
View Answer
Explanation: In order to compare two people are taken at a time.
Then, there would be 15C2 comparisons (Everyone is compared with the other).
15C2 = 105
6. How many three digit numbers can be formed from 2, 1, 5, 4, 8, 6 such that the digits are in descending order?
a) 18
b) 6! * 3!
c) 21
d) 20
View Answer
Explanation: Given that, 2, 1, 5, 4, 8 and 6.
Numbers starting with 86 – 4, 85 – 3, 84 – 2, 82 – 1 ⟹ 4 + 3 + 2 + 1 = 10 numbers.
Numbers starting with 65 – 3, 64 – 2, 62 – 1 ⟹ 3 + 2 + 1 = 7 numbers.
Numbers starting with 54 – 2, 52 – 1 ⟹ 2 + 1 = 3 numbers.
Numbers starting with 42 – 1 ⟹ 1 numbers.
Number of three digit numbers we can form = 10 + 7 + 3 + 1 = 21.
7. Three coins are tossed at a time. What is the probability of getting at least two tails?
a) 1 / 4
b) 3 / 4
c) 2 / 3
d) 1 / 2
View Answer
Explanation: When three coins are tossed, total number of outcomes will be (HHH, HHT, HTH, THH, TTT, TTH, THT, HTT) = 8.
Required type of outcomes = (TTT, TTH, HTT, THT).
Probability of getting at least two tails = 4 / 8 = \(\frac {1}{2}\).
8. How many ways can the word ‘PRACTICE’ be arranged, when repetition is allowed?
a) 8! * 2!
b) 88 / 2!
c) 88
d) 8! / 2!
View Answer
Explanation: The word ‘PRACTICE’ has 8 letters.
It is given that repetition is allowed.
In this the letter ‘C’ is repeated.
Let’s assume x = number of letters in the word. (When repetition is allowed we write it as xx).
y = (number of times the letter is repeated)!.
Number of ways the word can be arranged = xx / y! = 88 / 2!.
9. There are 5 children named M, N, O, P, Q who are made to sit on a bench. In how many ways they can be arranged if P and Q are always together?
a) 4!
b) 5! / 2!
c) 48
d) 24
View Answer
Explanation: Given, 5 children M, N, O, P and Q are made to sit on a bench.
It is also given that P and Q always sit together.
Consider M, N, O, PQ. We can arrange them in 4! Ways.
For every arrangement of PQ we will have a parallel arrangement with QP.
The arrangement can be done in 4! * 2! = 24 * 2 = 48ways.
10. Sunil has a pack of cards which contains 52 cards. He selects 3 cards at random, what will be the possibility of outcomes of the selected cards?
a) 21200
b) 24500
c) 25200
d) 22100
View Answer
Explanation: Given that, Sunil has a pack of cards.
He selects 3 cards at random.
The possible outcomes will be 52C3 (selecting 3 out of 52).
52C3 = 22100
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