What is the probability of sitting 5 girls and 5 boys alternatively in a row ?

The arragement of sitting of 5 Boys and 5 Girls alternatively in a row may start with either a Boy or a Girl. So 2 types of starting are possible.

`"Type I"->BGBGBGBGBG`

`"Typy II"->GBG BGBGBGB`

In each type 5 Boys and 5 Girls may take their positions in `5!` ways. So total number of possible arrangements becomes

`=2xx5!xx5!`

and this is the number of favourable events.

Again without restriction Boys and Girls may be arranged randomly in `10!` ways which is possible number of events.

So the probability of sitting in alternative manner will be given by
`P="favourable number of events"/"possible number of events"`

`=(2xx5!xx5!)/(10!)=1/126`

One precaution

Calculating the favourable number of events if we consider that Boys or Girls may take positions in 4 gaps in between two and 2 positions at two ends i.e. total 6 possible positions. then we arrive at an erroneous possible number of events as `6!xx5!`.

This is wrong.Because in this calculation 2 Boys or 2 Girls may come side by side in some of the arrangements, which is not desired here.