Question

A family of $4$ brothers and $3$ sisters is to be arranged in a row for a photograph. The number of ways in which they can seated if all the sisters are to sit together is:

Answer 1

Here this type of question is based on permutation and combination. Where we have to do an arrangement of 4 brothers and 3 sisters into a sequence or linear. We will make all sisters sit together so all sisters as one bundle so we can get $5$ elements to arrange.

Complete step-by-step answer:
It is given that there are $4$ brothers and $3$ sisters In a family.
According to the question we have to make all sisters sitting together.
For that, we count all sisters as one bundle.
Now we have a total of $5$ places to arrange them.
We can arrange $4$ brothers and $3$ sisters in $5!$ ways $= 120$ ways
But there are $3$ sisters we consider them as one bundle but we can arrange them inside.
We can arrange $3$ sisters in $3!$ ways $= 6$ ways.
Total arrangement of $4$ brothers and $3$ sisters in which all sisters are to sit together $= 6 \times 120$
$= 720$ ways
Therefore, the number of ways in which they can seat if all the sisters are to sit together is $= 720$

Note: A permutation of a set is, loosely speaking, an arrangement of its members into a sequence or linear order, or if the set is already ordered, a rearrangement of its elements. Whenever they ask us to sit together there we have to make one bundle of those elements. After arranging all elements together we have to rearrange those elements which are inside that one bundle.