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  • A group of 6 friends, including 3 siblings, is going to sit in a row for a photo. If the siblings must sit together, and no two siblings can sit next to each other, in ho

A group of 6 friends, including 3 siblings, is going to sit in a row for a photo. If the siblings must sit together, and no two siblings can sit next to each other, in how many different ways can they be arranged?

 

Option: 1

240


Option: 2

480


Option: 3

360


Option: 4

120


Answers (1)

best_answer

To calculate the number of different ways the group of 6 friends can be arranged in a row for a photo, with the siblings sitting together but no two siblings sitting next to each other, we can consider the following:

Let's label the siblings within the group as \mathrm{A, A^{\prime}, B}, and \mathrm{B^{\prime}}, where A and \mathrm{A^{\prime} }are the first pair of siblings, and B and B ' are the second pair of siblings.

Case 1: A and \mathrm{A^{\prime}} sit together.

In this case, we treat A and \mathrm{A^{\prime}} as a single entity. Therefore, we have 5 entities to arrange: the combined entity of A and \mathrm{A^{\prime}, B, B^{\prime},} and the remaining 2 friends.

The number of different ways to arrange these 5 entities in a row is 5 !.

Within the combined entity of A and \mathrm{A^{\prime}}, the siblings A and \mathrm{A^{\prime}} can be arranged among themselves, which gives us 2 ! possibilities.

Case 2: B and B' sit together.

This case is similar to Case 1 , so we have the same number of possibilities: 5 ! \times 2 !.

Therefore, the total number of different ways to arrange the siblings and the remaining friends is:

(5 ! \times 2 !)+(5 ! \times 2 !)=2 \times 5 ! \times 2 !=2 \times 120 \times 2=480 .

Therefore, there are 480 different ways the group of 6 friends can be arranged in a row for a photo, with the siblings sitting together but no two siblings sitting next to each other.

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Gaurav

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