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CAT Exam  >  CAT Test  >  Level-wise Tests  >  Test Level 2: Permutation & Combination - 1 - CAT MCQ

Test Level 2: Permutation & Combination - 1 CAT MCQs solutions - Free


MCQ Practice Test & Solutions: Test Level 2: Permutation & Combination - 1 (10 Questions)

You can prepare effectively for CAT Level-wise Tests for CAT with this dedicated MCQ Practice Test (available with solutions) on the important topic of "Test Level 2: Permutation & Combination - 1". These 10 questions have been designed by the experts with the latest curriculum of CAT 2026, to help you master the concept.

Test Highlights:

  • - Format: Multiple Choice Questions (MCQ)
  • - Duration: 20 minutes
  • - Number of Questions: 10

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Test Level 2: Permutation & Combination - 1 - Question 1
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In how many ways can 45 men be allotted to 3 different districts, if the districts are to be covered by 10, 15 and 20 men?

Detailed Solution: Question 1

The number of ways is the same as the number of ways in which m + n + p things can be divided into three groups containing m, n and p, which is 
∴ The required number of different allotments 

Test Level 2: Permutation & Combination - 1 - Question 2
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In the word 'MATHEMATICS', the positions of vowels and consonants are kept unchanged. How many different words can be formed by using different arrangements of the remaining letters?

Detailed Solution: Question 2


Different ways of arranging letters so that the positions of vowels and consonants remain unchanged 

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Test Level 2: Permutation & Combination - 1 - Question 3
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In how many ways can 8 identical balls be distributed in 3 distinct boxes so that none of the boxes remains empty?

Detailed Solution: Question 3

Suppose there are 3 boxes B1, B2, B3
In total there are 8 boxes we are going to put 1 ball in each box
So, the remaining number of balls after putting 1 ball in each box = 8 - 3 = 5
so number of remaining balls n = 5
the number of boxes r = 3
The number of ways for arranging the balls in boxes without any condition = (n + r - 1)Cr-1
n + r - 1 = 5 + 3 - 1 = 7
r - 1 = 3 - 1 = 2

Test Level 2: Permutation & Combination - 1 - Question 4
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How many words can be formed with the letters of the word PATLIPUTRA without changing the relative positions of vowels and consonants?  

Detailed Solution: Question 4


AIUA -- Vowels and 'PTLPTR' -- Consonants
Number of words that can be formed without changing the relative positions of vowels and consonants

= 2160

Test Level 2: Permutation & Combination - 1 - Question 5
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In how many ways can two dice be rolled, so that the sum of the values on the upper faces is divisible by three?  

Detailed Solution: Question 5

Out of the 36 possible outcomes, 12 outcomes will yield a score divisible by 3, i.e. a score of either 3, 6, 9 or 12.
The 12 outcomes will be various arrangements of (1, 2), (2, 1), (2, 4), (4, 2), (3, 3), (3, 6), (6, 3), (4, 5), (5, 4), (5, 1),(1, 5) and (6, 6).

Test Level 2: Permutation & Combination - 1 - Question 6
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A store carries four styles of pants. For each type, there are ten different possible waist sizes, six different pant lengths and four colour choices. How many different types of pants could the store have?  

Detailed Solution: Question 6

Using the fundamental principle of multiplication, total number of types = 10 × 6 × 4 × 4 = 960

Test Level 2: Permutation & Combination - 1 - Question 7
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What is the number of words that can be formed by using the letters A, B, C, D, E and F, taken three at a time, if each word contains at least one vowel?

Detailed Solution: Question 7

Since we are taking three letters at a time, the total number of three letter words is 6P4 = 6 × 5 × 4 = 120. The number of words not containing A and E is 4P3 = 4 × 3 × 2 = 24. Therefore, the required number of words is 96.

Test Level 2: Permutation & Combination - 1 - Question 8
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In how many ways can 5 different coins be placed in small squares of a chess board (8 × 8), such that no two coins should be in the same row or column?

Detailed Solution: Question 8

The first coin can be placed in any box out of 64 boxes.
Second coin cannot be in the same row or column. So, there should be 64 - (8 + 7) = 49 squares left, and so on.
Required number of ways = 64 × 49 × 36 × 25 × 16 = (6720)2

Test Level 2: Permutation & Combination - 1 - Question 9
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A group consists of 6 men and 8 women. A committee of 4 has to be formed with at least one man. The probability that the committee so formed has a greater number of men than women is

Detailed Solution: Question 9

There are 6 men and 8 women.

Required number of ways (Number of men > Number of women) = C(6, 3) x C(8, 1) + C(6, 4)
So, the required probability 


Hence, option (1) is the correct answer.

Test Level 2: Permutation & Combination - 1 - Question 10
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8 men and 6 women are working in an office. 9 of them are to be chosen for a meeting that should include at least 3 men and 2 women. 2 men present in the last meeting must attend this meeting as well. The number of ways in which they can be selected is

Detailed Solution: Question 10

There are 8 men and 6 women. Two men are fixed, so we have 6 men and 6 women for selection. Now, out of 9 persons, 2 men are fixed, so we have to select the rest 7 persons, including at least 1 man and 2 women.

Possible cases are as follows.
(1) 1 man, 6 women = C(6, 1) × C(6, 6)
(2) 2 men, 5 women = C(6, 2) × C(6, 5)
(3) 3 men, 4 women = C(6, 3) × C(6, 4)
(4) 4 men, 3 women = C(6, 4) × C(6, 3)
(5) 5 men, 2 women = C(6, 5) × C(6, 2)
So, total number of ways = [C(6, 1) × C(6, 6) + C(6, 2) × C(6, 5) + C(6, 3) × C(6, 4) + C(6, 4) × C(6, 3) + C(6, 5) × C(6, 2)]

= 6 + 90 + 300 + 300 + 90 = 786

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About this CAT Practice Test

This test contains 10 MCQs on Test Level 2: Permutation & Combination - 1 with detailed solutions for each question. It is part of the Level-wise Tests for CAT course on EduRev, designed to align with the current CAT syllabus. Each solution explains not just the correct answer but also gives you an understanding of the topic — not just to attempt the question but also help you prepare for the exam with practice.

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