Test: Permutation & Combination- 2

# Test: Permutation & Combination- 2

Test Description

## 10 Questions MCQ Test CSAT Preparation for UPSC CSE | Test: Permutation & Combination- 2

Test: Permutation & Combination- 2 for Banking Exams 2023 is part of CSAT Preparation for UPSC CSE preparation. The Test: Permutation & Combination- 2 questions and answers have been prepared according to the Banking Exams exam syllabus.The Test: Permutation & Combination- 2 MCQs are made for Banking Exams 2023 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Permutation & Combination- 2 below.
Solutions of Test: Permutation & Combination- 2 questions in English are available as part of our CSAT Preparation for UPSC CSE for Banking Exams & Test: Permutation & Combination- 2 solutions in Hindi for CSAT Preparation for UPSC CSE course. Download more important topics, notes, lectures and mock test series for Banking Exams Exam by signing up for free. Attempt Test: Permutation & Combination- 2 | 10 questions in 10 minutes | Mock test for Banking Exams preparation | Free important questions MCQ to study CSAT Preparation for UPSC CSE for Banking Exams Exam | Download free PDF with solutions
 1 Crore+ students have signed up on EduRev. Have you?
Test: Permutation & Combination- 2 - Question 1

### In a group of 6 boys and 4 girls, four children are to be selected. In how many different ways can they be selected such that at least one boy should be there?

Detailed Solution for Test: Permutation & Combination- 2 - Question 1

The total number of ways to select 4 children out of 10 total children is (104)=10∗9∗8∗74∗3∗2∗1=210. However, because at least one boy must be in the group, we have to subtract 1 to account for the case where all four children are girls (There is (44)=1 such case). The answer is 210−1=209.

Test: Permutation & Combination- 2 - Question 2

### MCQ (Multiple Choice Questions) with solution are available for Practice, which would help you prepare for Permutation and Combination under Quantitative Aptitude. You can practice these practice quizzes as per your speed and improvise the topic. The same topic is covered under various competitive examinations like - CAT, GMAT, Bank PO, SSC and other competitive examinations. Q. Out of 7 consonants and 4 vowels, how many words of 3 consonants and 2 vowels can be formed?

Detailed Solution for Test: Permutation & Combination- 2 - Question 2

Number of groups, each having 3 consonants and 2 vowels = 210. Required number of ways = (210 x 120) = 25200.

Test: Permutation & Combination- 2 - Question 3

### From a group of 7 men and 6 women, five persons are to be selected to form a committee so that at least 3 men are there on the committee. In how many ways can it be done?

Detailed Solution for Test: Permutation & Combination- 2 - Question 3

We have to consider the 3 possible ways to accomplish this :
5 men & 0 women
4 men & 1 women
3 men & 2 women

Hence the required number is
C(7,5).C(6,0) + C(7,4).C(6,1) + C(7,3).C(6,2)
=21x1 + 35x6 + 35x15
=21+210+525 = 756

Test: Permutation & Combination- 2 - Question 4

In how many ways can a group of 5 men and 2 women be made out of a total of 7 men and 3 women?

Detailed Solution for Test: Permutation & Combination- 2 - Question 4

5 men can be chosen out of 7 men in 7C5 = 7C2 = 21 ways.

2 women out of 3 women in 3C2 = 3C1 = 3 ways.

Hence by fundamental counting principle, group of 5 men and 2 women can be chosen from 7 men and 3 women in (21)(3) = 63 ways

Test: Permutation & Combination- 2 - Question 5

In how many different ways can the letters of the word 'LEADING' be arranged such that the vowels should always come together?

Detailed Solution for Test: Permutation & Combination- 2 - Question 5

The word 'LEADING' has 7 different letters.

When the vowels EAI are always together, they can be supposed to form one letter.

Then, we have to arrange the letters LNDG (EAI).

Now, 5 (4 + 1) letters can be arranged in 5! = 120 ways.

The vowels (EAI) can be arranged among themselves in 3! = 6 ways.

Required number of ways = (120 x 6) = 720.

Test: Permutation & Combination- 2 - Question 6

In how many different ways can the letters of the word 'CORPORATION' be arranged so that the vowels always come together?

Detailed Solution for Test: Permutation & Combination- 2 - Question 6

Vowels in the word "CORPORATION" are O,O,A,I,O
Lets make it as CRPRTN(OOAIO)
This has 7 lettes, where R is twice so value = 7!/2!
= 2520
Vowel O is 3 times, so vowels can be arranged = 5!/3!
= 20
Total number of words = 2520 * 20 = 50400

Test: Permutation & Combination- 2 - Question 7

In how many different ways can the letters of the word 'MATHEMATICS' be arranged such that the vowels must always come together?

Detailed Solution for Test: Permutation & Combination- 2 - Question 7

In the word 'MATHEMATICS', we'll consider all the vowels AEAI together as one letter.
Thus, we have MTHMTCS (AEAI).
Now, we have to arrange 8 letters, out of which M occurs twice, T occurs twice
Number of ways of arranging these letters =8! / ((2!)(2!))= 10080.

Now, AEAI has 4 letters in which A occurs 2 times and the rest are different.
Number of ways of arranging these letters =4! / 2!= 12.

Required number of words = (10080 x 12) = 120960

Test: Permutation & Combination- 2 - Question 8

In how many different ways can the letters of the word 'OPTICAL' be arranged so that the vowels always come together?

Detailed Solution for Test: Permutation & Combination- 2 - Question 8

The word 'OPTICAL' contains 7 different letters. When the vowels OIA are always together, they can be supposed to form one letter. Then, we have to arrange the letters PTCL (OIA). Now, 5 letters can be arranged in 5!=120 ways. The vowels (OIA) can be arranged among themselves in 3!=6 ways. Required number of ways =(120∗6)=720.

Test: Permutation & Combination- 2 - Question 9

There are 8 men and 10 women and you need to form a committee of 5 men and 6 women. In how many ways can the committee be formed?

Detailed Solution for Test: Permutation & Combination- 2 - Question 9

Required number of ways= (8C5 x 10C6)
= (8C3 x 10C4)
= (8 x 7 x 6 /3 x 2 x 1)  x  (10 x 9 x 8 x 7 / 4 x 3 x 2 x 1)
= 11760.

Test: Permutation & Combination- 2 - Question 10

How many 3-letter words with or without meaning, can be formed out of the letters of the word, 'LOGARITHMS', if repetition of letters is not allowed?

Detailed Solution for Test: Permutation & Combination- 2 - Question 10

The word 'LOGARITHMS' has 10 different letters.
Hence, the number of 3-letter words(with or without meaning) formed by using these letters
10P3
= 10 x 9 x 8
= 720

## CSAT Preparation for UPSC CSE

72 videos|64 docs|92 tests
 Use Code STAYHOME200 and get INR 200 additional OFF Use Coupon Code
Information about Test: Permutation & Combination- 2 Page
In this test you can find the Exam questions for Test: Permutation & Combination- 2 solved & explained in the simplest way possible. Besides giving Questions and answers for Test: Permutation & Combination- 2, EduRev gives you an ample number of Online tests for practice

## CSAT Preparation for UPSC CSE

72 videos|64 docs|92 tests