Question

Question 24 :
In how many ways 5 men (M1, M2, M3,
M4, M5) can be seated in a line such
that M1 and M2 are always seated
together?
5!
4!
4! 2!
5! 2!​

Answer 1

Answer:

4!

Step-by-step explanation:

Since, M1 & M2 should be seated together we consider both of men as one. so now there are total 4.

Since, there are 4 different men taken 1 at a time

we have,

4 × 3 × 2 × 1 = 4!

Answer 2

In (4! 2! ways) 5 men can be seated in a line in such a way that M1 and M2 are always seated together. Option (c) is correct.

Given:

5 men (M1, M2, M3, M4, M5) can be seated in a line such that M1 and M2 are always seated together.

To find:

The number of ways 5 men can be seated with the given condition needs to be determined.

Solution:

There are 5 men in which M1 and M2 are always seated together. So, treating them as one unit.

So, the remaining is M3, M4, and M5.

Using permutation and combination, these can be arranged as M3, M4, M5, and (M1M2) = 4! ways = 4×3×2= 24.

Also, two men (M1M2) can also be arranged in 2! = 2×1 = 2 ways.

The number of ways 5 men can be seated so that M1 and M2 are always seated together will be 24×2 = 48 ways = 4! ×2! ways.

To learn more about permutation and combination visit:

https://brainly.in/question/9087739?referrer=searchResults

https://brainly.in/question/13812790?referrer=searchResults

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