Permutations quetion , slight doubt
THe question is
THe number of ways in whcih 4 distinct balls can be put in 4boxes labelled a,b,c,d so that exactly one box remains empty:
So what I have done till now is as follows:
1. Case 1: Box "a" remains empty. So we have 4balls which need to go in 3boxes so that every box should have atleast 1ball.
So, we select 3balls from 4 to fill in one ball in each box. (4c3 * 3! ) and the remaining ball can go in any of the b,c,d boxes.
So , total cases where Box "a" is remaining empty are 4C3 * 3! * 3 = 72
BY similarity , there will be four cases in total. So the answer is 4 * 4C3 * 3! * 3 = 288 .
But the correct answer is 144. That means I am doubling up some number which I am unable to figure out.