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

Mar 8, 2017

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.

$\text{Type I} \to B G B G B G B G B G$

$\text{Typy II} \to G B G B G B G B G B$

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
$P = \text{favourable number of events"/"possible number of events}$

=(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.