Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • A group of 7 friends, including 2 pairs of siblings, is going to sit in a row for a photo. If the siblings must sit together, but the two pairs cannot sit next to each ot

A group of 7 friends, including 2 pairs of siblings, is going to sit in a row for a photo. If the siblings must sit together, but the two pairs cannot sit next to each other, in how many different ways can they be arranged?

 

Option: 1

7632

 


Option: 2

8225

 


Option: 3

4552

 


Option: 4

6895


Answers (1)

best_answer

To calculate the number of different ways the group of 7 friends can be arranged in a row for a photo, with the siblings sitting together but the two pairs not sitting next to each other, we can consider the following:

Let's label the siblings within each pair 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 \mathrm{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 6 entities to arrange: the combined entity of A and \mathrm{A^{\prime}, B, B^{\prime}}, and the remaining 3 friends.

The number of different ways to arrange these 6 entities in a row is 6 !.

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: 6 ! \times 2 !.

Case 3: Neither pair sits together.

In this case, we need to consider the arrangement of A and \mathrm{A^{\prime}, B} and \mathrm{B^{\prime}}, and the remaining 3 friends.

The number of different ways to arrange these 7 entities in a row is 7 !.

Within the combined entity of A and A', the siblings A and A' can be arranged among themselves, which gives us 2! possibilities.

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

Therefore, the total number of different ways to arrange the group of 7 friends, with the siblings sitting together but the two pairs not sitting next to each other, is:

(6 ! \times 2 !)+(6 ! \times 2 !)+(7 ! \times 2 ! \times 2 !)=(6 ! \times 4)+(7 ! \times 4)=2,592+5,040=7,632 .

Therefore, there are 7,632 different ways the group of 7 friends can be arranged in a row for a photo, with the siblings sitting together but the two pairs not sitting next to each other.

Posted by

Ritika Kankaria

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Latest Question