How many ways are there to put $22$ identical balls into $5$ boxes, with each box receiving at least $2$ balls?

Question

How many ways are there to put $22$ identical balls into $5$ boxes, with each box receiving at least $2$ balls?

The answer is $16$ choose $4$, i.e., $_{16}C_4$, but can someone explain it to me?

Answer 1

This is a classic stars and bars problem. The solution you have presumes that you must put at least two balls in each box. Is that correct?

Answer 2

As we have at least two balls in each box, we know where 10 of the balls will be, so the question is equivalent to asking how many ways (up to reordering) we can put 12 balls in 5 boxes. This is the same as asking how many ways we can express 12 as a sum of 5 non-negative integers. There are 45 possible ways of doing this:

• 12 = 12+0+0+0+0
• 12 = 11+1+0+0+0
• 12 = 10+2+0+0+0
• 12 = 10+1+1+0+0
• 12 = 9+3+0+0+0
• 12 = 9+2+1+0+0
• 12 = 9+1+1+1+0
• 12 = 8+4+0+0+0
• 12 = 8+3+1+0+0
• 12 = 8+2+1+1+0
• 12 = 8+1+1+1+1
• 12 = 7+5+0+0+0
• 12 = 7+4+1+0+0
• 12 = 7+3+2+0+0
• 12 = 7+3+1+1+0
• 12 = 7+2+2+1+0
• 12 = 7+2+1+1+1
• 12 = 6+6+0+0+0
• 12 = 6+5+1+0+0
• 12 = 6+4+2+0+0
• 12 = 6+4+1+1+0
• 12 = 6+3+3+0+0
• 12 = 6+3+2+1+0
• 12 = 6+3+1+1+1
• 12 = 6+2+2+2+0
• 12 = 6+2+2+1+1
• 12 = 5+5+2+0+0
• 12 = 5+5+1+1+0
• 12 = 5+4+3+0+0
• 12 = 5+4+2+1+0
• 12 = 5+4+1+1+1
• 12 = 5+3+3+1+0
• 12 = 5+3+2+2+0
• 12 = 5+3+2+1+1
• 12 = 5+2+2+2+1
• 12 = 4+4+4+0+0
• 12 = 4+4+3+1+0
• 12 = 4+4+2+2+2
• 12 = 4+3+3+2+0
• 12 = 4+3+3+1+1
• 12 = 4+3+2+2+1
• 12 = 4+2+2+2+2
• 12 = 3+3+3+3+0
• 12 = 3+3+3+2+1
• 12 = 3+3+2+2+2

So there are 45 possible ways of placing 12 identical balls into 5 identical boxes, and hence 45 possible ways of placing 22 identical balls in 5 boxes such that every box contains at least 2 balls.