Combination of 5 balls

Question

The total number of ways in which 5 balls of $different$ colours can be distributed among 3 $persons$ so that each person gets at least one ball.

This is a case of distinct to distinct.

Case 1. groups of 1, 1, 3 gives ${5 \choose 1}$ ${4 \choose 1}$ ${3\choose 3}$= 20

Case 2 groups of 1, 2, 2 gives ${5 \choose 1}$ ${4 \choose 2}$ ${2 \choose 2}$ = 30

Now here my problem arises.

Take case 1

Assume balls to be A B C D E. You make groups A, B, CDE.

Let persons be 1 2 3.

The following distributions can be done (one column is one distribution)

PERSON 1   A        A      B       B      CDE       CDE

PERSON 2   B       CDE     A      CDE      A         B

PERSON 3  CDE       B      CDE      A      B         A

That is in case 1, each group made can be distributed in 6 ways (3!). Therefore 20 groups can be distributed in 20 x 6 = 120 ways.

Similarly for case 2 we get 30 x 6 = 180 ways.

Therefore total is equal to 600 ways.

But the answer given is 150. Each distribution is taken as 3!/2!. Why is that ? The question specifies different colour balls and to person and persons are always different.

Answer 1

Hint: if you simply multiply with $3!$, then $6$ permutation of $A,B,CDE$ are different from $6$ permutation of $B,A,CDE$

similar logic to $1,2,2$

A neat solution is using PIE: $3^{5}-3\times 2^{5}+3\times 1^{5}=150$

Answer 2

You can distribute 5 balls with at least one ball to each person in 6 ways with the distribution as follows: (1,1,3), (1,2,2),(2,1,2),(2,2,1),(1,3,1),(3,1,1) In case 1, person 1 gets 1 ball, person 2 gets 1 ball and person 3 gets the remaining. In case 2, person 1 gets 1 ball, person 2 gets 2 and person 3 gets remaining 2. I hope you can understand the rest of the cases also. Now, case 1 can happen in $ {5 \choose 1} ×{ 4\choose 1} × {3\choose 3 }= 20 $ ways. Case 2 can happen in${ 5 \choose 1}× {4 \choose 2}× {2\choose 2} = 30 $ ways.
Similarly, by applying suitable selection formula for other cases with the help of the distribution given, case 3 gives $30$, case 5 gives $30$, case 5 gives $20$ and case 6 gives$ 20$ ways. Summing up, we get 150.