Find the number of ways in which 4 boys and 3 girls can be seated in a row so that all girls are together ?

Concept:

The number of ways to arrange n distinct things taken all at a time is given by: \({\;^n}{P_n} = n!\)

Let r and n be the positive integers such that, 0 ≤ r ≤ n. Then the number of ways to arrange r thing taken at a time out of n different things is given by: \({\;^n}{P_r} = \frac{{n!}}{{\left( {n - r} \right)!}}\)

Calculation:

Given: There are 4 boys and 3 girls that can be seated in a row such that all girls are together.

First let's calculate number of ways to arrange boys in a row.

∵ There are 4 boys so they will occupy 4 places in \({\;^4}{P_4} = 4!\) ways = 24 ways

We can arrange the all-girls together total 5 positions shown by cross mark: {X X X} B B B B

Number of ways in which all girls are together but interchange = \({\;^5}{P_5} × 3!\)

As we know that, \({\;^n}{P_r} = \frac{{n!}}{{\left( {n - r} \right)!}}\)

⇒ 5! × 3!

⇒ 120 × 6

∴ The required number of ways = 120 × 6 = 720