# Permutation problem 13 Nov

**Question**

!!16!! people, !!8!! chairs on each side of the table . Four men sit on one particular side and two on other side. Number of ways can they be seated?

!!a)" "(8!8!10!)/(2!4!)!!

!!b)" "(8!8!10!)/(4!6!)!!

!!c)" "(8!8!10!)/(2!6!)!!

!!a)" "(8!8!10!)/(2!4!6!)!!

**Solution**

In the given figure we can see that there are 16 people and !!8!! chairs are arranged on each side of the table.

Since !!4!! men sit on one particular side, hence number of ways they can be seated = !!(8c_4)xx4!!!

!!2!! men sit on other side of the table, hence number of ways they can be seated = !!(8c_2)xx2!!!

Rest other can seat !!10!!! ways.

Now total number of ways !!16!! men can be seated = !!((8c_4)xx4!)xx((8c_2)xx2!)xx10!!!

!!= (8!4!)/(4!4!)xx(8!2!)/(2!6!)xx10!!!

!!= (8!)/(4!)xx(8!)/(6!)xx10!!!

!!= (8!8!10!)/(4!6!)!!