Permutations and Combinations: Distributing Balls Among People

[erisedk]

Homework Statement

The total number of ways in which 5 balls of different colors can be distributed among 3 persons so that each person gets at least one ball is
Ans: 150

Homework Equations

The Attempt at a Solution

I don't understand what's wrong with my answer.
In case of each person getting one ball and the remaining two balls going to a single person, I have 5*4*3*3 ways. 5*4*3 for the first three balls and then the remaining two balls have three options, therefore the final *3

In case of the remaining two balls going to different people, I have 5*4*3*6 ways to do so, the final*6 because I have 3 spots and I have two pick two balls, and order matters as the balls are different.

Adding 5*4*3*3 + 5*4*3*6, I get 540.

What am I doing wrong?

 

[mfb]

50 is wrong, but 540 is not right either. You also count the order the persons get the balls as different. If the colors are ABCDE and the persons are 1, 2, 3, you count "A to 1, B to 2, C to 3 and then D and E to 1" as different from "D to 1, B to 2, C to 3 and then A and E to 1".

Better start picking the person to get 3 balls, and then look at the other two persons. The other part has the same issue.

 

[vela]

Another way to look the 1-1-3 distribution is is to count the number of ways you can give two of the three people one ball each. The third person ends up with the rest.

 

[haruspex]

You also count the order the persons get the balls as different.

erisedk, just in case that's not clear, mfb is saying you mistakenly counted those orders as different, not that you should count them as different.

 

[erisedk]

Oh, ok.
I get it now.
Thanks :)