Ace booklet

Question

How many ways $4$ boys and $4$ girls can sit in a row so that no two girl are sitting side by side

Answer 1

There are $4$ Boys & $4$ Girls

Criteria given is "All the Boys & Girls will be seat in a row" & "No two girls will be seat together"

I'll make Boys seat first in the row.

So, Boys can be seat in $4$ seats in $4!$ ways.

Now, there are $5$ seats left vacant.

Now, from these $5$ vacant seats, we could choose $4$ seats & make Girls seat.

& This can be done in $^5C_4 \times 4!$ or we can say $^5P_4$ ways

[ Why $^5C_4 \times 4!$ $ \rightarrow$ $^5C_4$ because we'll choose $4$ vacant seats out of $5$ seats & $4!$ because after choosing $4$ seats, we can arrange them in $4$! ways  And $^5C_4 \times 4!$ can be written as$^5P_4$  ]

∴ Total number of ways will be = $4! \times ^5C_4 \times 4! $

$\qquad \qquad \qquad \qquad \qquad \qquad =4! \times ^5P_4 $

$\qquad \qquad \qquad \qquad \qquad \qquad = 24 \times 120$

$\qquad \qquad \qquad \qquad \qquad \qquad = 2880$

Answer 2

Firstly boys should be sit like     B1   B2 B3 B4 ,SO they can sit in 4! Way and now between the boys put girls,,so there are 5 gaps between boys girls can be arrange in $P_{4}^{5} $,

So answer will be 4!×$P_{4}^{5}$