Probability With Buckets And Balls

Question

There are 5 non identical balls and 5 non identical buckets. You can place any amount of balls in a single bucket.

a) In how many cases are there precisely one bucket empty? b) In how many cases are there precisely three empty buckets?

Answer 1

Part A

First take the first bucket empty, then number of ways is only four (of choosing number of balls in bucket-not arranging [as all balls are distinct]) as other $4$ buckets needs to have atleast one ball, so no of balls would be: $1$ ball in each of $3$ buckets and $2$ balls in $4^{th}$.Now empty bucket can be chosen is $5$ ways.Now we can arrange all $5$ balls in $5!$ ways and in the bucket containing $2$ balls, the order doesn't matter so we divide by $2$.So putting balls inside would be: $$\large W=\underbrace{4}_{\text{no. of balls}}\times\underbrace{5}_{\text{empty bucket}}\times\underbrace{\frac{5!}2}_{\text{arrangement}}=1200$$

Part B

Working similiarly, taking balls in buckets to be $(1,4),(2,3)$, each arrangement twice since all buckets are distinct $$\large W=\underbrace{2}_{\text{reflection of (a,b) to (b,a)}}\times\underbrace{5!}_{\text{arrangement}}\times\underbrace{\binom53}_{\text{choosing 3 empty buckets}}\times\left\{\underbrace{\frac1{4!1}}_{(1,4) and (4,1)}+\underbrace{\frac1{3!2!}}_{(2,3) and (3,2)}\right\}=300$$