In how many ways can you allocate 6 boys to 4 different rooms if no room can be empty?
1. 1020
2 540
3. 480
4. 1680
1. 1020
2 540
3. 480
4. 1680
These are the possibilities : (3,1,1,1) and (2,2,1,1) => Means EITHER one room can have 3 boys and the rest will have one each OR two rooms will have 2 boys each and the other two rooms will have 1 boy eachsrcc25anu wrote:In how many ways can you allocate 6 boys to 4 different rooms if no room can be empty?
1. 1020
2 540
3. 480
4. 1680
So, total no. of ways = (No. of ways to select one room which will have 3 boys) * (No. of ways to select the boys for the 4 rooms) + (No. of ways to select two rooms which will have 2 boys each ) * (No. of ways to select the boys for the 4 rooms)
= (4C1)*(6C3*3*2*1) + (4C2)*(6C2*4C2/2! * 2 * 1) [We divide 6C2*4C2 by 2!, since the order of the two rooms which will have 2 boys, doesn't matter]
= 4*20*3*2 + 6* (5*6/2 * 3*4/2)/2 * 2* 1 = 480 + 6* 45*2 = 480+540 = 1020 A
