Test: Permutation of Distinct Objects - Commerce MCQ

# Test: Permutation of Distinct Objects - Commerce MCQ

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## 10 Questions MCQ Test - Test: Permutation of Distinct Objects

Test: Permutation of Distinct Objects for Commerce 2024 is part of Commerce preparation. The Test: Permutation of Distinct Objects questions and answers have been prepared according to the Commerce exam syllabus.The Test: Permutation of Distinct Objects MCQs are made for Commerce 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Permutation of Distinct Objects below.
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Test: Permutation of Distinct Objects - Question 1

### In how many ways 4 boys and 3 girls can be seated in a row so that they are alternate.

Detailed Solution for Test: Permutation of Distinct Objects - Question 1

To solve this problem, we need to find the number of ways to seat 4 boys and 3 girls in a row such that boys and girls alternate.

Step 1: Determine the possible seating arrangements

Since there are 4 boys and 3 girls, one possible alternating pattern would be: B G B G B G B (boys first) or G B G B G B G (girls first).

Step 2: Arrange boys and girls separately

• Arrange the boys: There are 4 boys, so they can be arranged in 4! ways.
• Arrange the girls: There are 3 girls, so they can be arranged in 3! ways.

Step 3: Calculate total arrangements

• For the B G B G B G B pattern:

• Number of arrangements = 4!×3!
4!=24 and 3!=6

Step 4: Consider both patterns

Since we can start with either a boy or a girl, and each case is independent, we multiply the number of arrangements by 2:

Total number of arrangements=144×2=288

The number of ways to arrange 4 boys and 3 girls alternately is 288.

2. 288

Test: Permutation of Distinct Objects - Question 2

### In how many ways 2 directors and 3 executives can be arranged for a meeting? If there are 6 chairs available two on one side and remaining four on the other side of the table and the two directors has to be together on one side and the executives on the other side.

Detailed Solution for Test: Permutation of Distinct Objects - Question 2

3 executives in 4 chairs = 4P3 ways
Total no. of ways = 2! * 4P3
=> 2*1*(4!/1!)
= 2*24
= 48

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Test: Permutation of Distinct Objects - Question 3

### A lady arranges a dinner party for 6 guests .The number of ways in which they may be selected from among 10 friends if 2 of the friends will not attend the party together is

Detailed Solution for Test: Permutation of Distinct Objects - Question 3

Let the friends be A,B,C,D,E,F,G,H,I,J and assume A and B will not battend together
Case 1 : Both of them will not attend the party : Now we have to select the 6 guests from the remaing 8 members
Then no of ways is 8C6 = 28 ways.
Case 2 : Either of them are selected for the party :Now we have to select the 5 guests from the remaing 8 members and  one from A and B
Then the no of ways = 2C1 x 8C5 =112 ways.
Therefore total number of ways is 28+ 112 = 140 ways,

Test: Permutation of Distinct Objects - Question 4

What is the value of npo

Detailed Solution for Test: Permutation of Distinct Objects - Question 4

nP0
= n!/(n-0)!
=> n!/n!  = 1

Test: Permutation of Distinct Objects - Question 5

Find the value of n if nP1 = 10

Detailed Solution for Test: Permutation of Distinct Objects - Question 5

nPr = 10
nPr = n!/(n-r)!
10 = n!/(n-1)!
10 = [n!.(n-1)!]/(n-1)!
n! = 10

Test: Permutation of Distinct Objects - Question 6

Number of signals that can be made using 2 flags out of given 4 flags.

Detailed Solution for Test: Permutation of Distinct Objects - Question 6

No. of ways of selecting two flags out of four = 4C2
So, total possible different signals generated =  4C2×2!
⟹ 6×2=12

Test: Permutation of Distinct Objects - Question 7

What is the value of npn

Detailed Solution for Test: Permutation of Distinct Objects - Question 7

nPn
= n!/(n-n)!
= n!

Test: Permutation of Distinct Objects - Question 8

Detailed Solution for Test: Permutation of Distinct Objects - Question 8
7! /5! ×3!
=7×6×5! /5! ×3×2×1
=7
Test: Permutation of Distinct Objects - Question 9

5 singers are to be given performance in a concert. How many ways can they be arranged in a order

Detailed Solution for Test: Permutation of Distinct Objects - Question 9

5P5  ⇒ 5!/(5-5)!
= 5!/0!
= 5!/1 ⇒ 5!
= 120 ways

Test: Permutation of Distinct Objects - Question 10

What is the value of 0!

Detailed Solution for Test: Permutation of Distinct Objects - Question 10

The value of 0! = 1

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