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.