Question

in how many ways can $5$ balls be placed in $4$ boxes if the balls are distinguishable, and the boxes are indistinguishable?

Answer 1

For placing 5 distinguishable balls into 4 indistinguishable boxes, consider the partitions of the number 5. There are 76 ways to do this, taking into account the various combinations for each partition and ensuring that the boxes are considered indistinguishable.

Step-by-step explanation:

The problem of placing distinguishable balls into indistinguishable boxes is a classic combinatorial problem. To solve the problem of how many ways 5 balls can be placed into 4 boxes, we can use the concept of distributions based on the partitions of the number 5 (the number of balls), since the boxes are indistinguishable.

For five distinguishable balls (labeled a, b, c, d, e), the partitions of 5 are:

• 5 (all in one box)
• 4+1 (four in one box, one in another)
• 3+2 (three in one box, two in another)
• 3+1+1 (three in one box, each of the others in its own box)
• 2+2+1 (two pairs in two boxes and the final one in its own box)
• 2+1+1+1 (two in one box and the others each in their own box)
• 1+1+1+1+1 (each in its own box)

We then evaluate each partition:

1. One box: Only 1 way, since all balls go into a single box.
2. 4+1: This can happen in 5 ways (the lone ball can be any of the five).
3. 3+2: This can happen in 10 ways ([a, b, c] in one box and [d, e] in another, for example).
4. 3+1+1: This can happen in 10 ways (three balls together can be chosen in 5C3 ways, and rest is fixed).
5. 2+2+1: This can happen in 30 ways (two pairs of balls can be chosen in 5C2*3C2/2 ways since pairs are indistinguishable, and the last ball is fixed).
6. 2+1+1+1: This can happen in 20 ways (the pair can be any of 5C2, and the rest is fixed).
7. No need to consider 1+1+1+1+1 since there are only 4 boxes.

Adding these up, there are 1 + 5 + 10 + 10 + 30 + 20 = 76 ways to place 5 distinguishable balls into 4 indistinguishable boxes.