Find the number of ways in which 12 children can be divided into 2 groups of 6 if two particular boys must be in different groups.
The number of ways of choosing 6 boys from the 12 to form the two groups of 6 boys is given by C(12,6)
In the case where the two boys are in one group, the number of ways of forming the groups is C(10, 4). This is due to the fact that 2 boys are taken together for a group, so now we need to choose another 4 boys from the 10 remaining to complete the group.
So the number of ways of forming the groups where the two boys are in different groups can be calculated by subtracting the case where they are together from the general case: this is given by C(12, 6) - C(10 , 6)
= 12! / (6!* 6!) - 10! / (6!*4!)
= 924 - 210
Therefore the number of ways the children can be divided is 714.
The number of ways of forming 6 boys 0f two groups in which two patcular boys must be in dofferent groups.
We know that when we are doing 6 boys of two groups it sufficient if we do the combination 6 of one group. The remaining unselected 6 boys is automatcally another group of six.
Since we want two particular boys in different groups of six, we identify them and keep them separate. From the rest of the 10 boys we can select 5 . After selecting 5 we add one of the separately kept 2 boys to the the selected group. The other boy automatically will be among the unseled boys. Thus we do this activity in 10C5 Or C(10 , 5)* 2 ways.
So the number of possible ways of grouping of 6 out of 12 with 2 particualar boys in separate groups is 10C5 = (10*9*8*7*6/5!)*2 = 504.