How many ways are there to put 6 balls in 3 boxes if the balls are distinguishable but the boxes are not? hint(s): the following cases are the most likely to be miscounted: * putting 2 balls into each box. * putting 3 balls into two boxes. * putting 4 balls into a box and 1 ball into the other 2 boxes. the answer is not 207.

This is essentially asking how many different ways to partition 6 into 3 segments.

I am assuming "no ball in a box" is not allowed.

6 can be partitioned as

(4,1,1), (3,2,1), and (2,2,2)

So, calculate each partition, we get

(6 choose 4) + (6 choose 3)*(3 choose 2) + (6 choose 2) * (4 choose 2)

= 15 + 20*3 + 15*6

=165

no he doesnt because "-16' is negatiive

step-by-step explanation: