# Find number of ways to put 15 medals into boxes of capacity 3, 5 and 8.

There are 15 different medals in a drawer. Suppose the medals in the drawer are put into three boxes that can hold at most 3, 5 and 8 medals respectively. Find the number of ways of distributing the medals.

My attempt:

The total capacity is 16, therefore there are 3 possible arrangements, where 1 missing in each box.

arrangement 2,5,8 + arrangement 3,4,8 + arrangement 3,5,7

$$= C_2^{15} \cdot C_5^{13} + C_3^{15} \cdot C_4^{12} + C_3^{15} \cdot C_5^{12}$$

$$= 720720$$

But the answer says $$3243240$$

• Your work seems correct to me - does the answer key come with an explanation of their solution? Commented Jul 14 at 5:23
• The answer does not come with explanation =( Commented Jul 14 at 6:15
• Are the medals distinct or identical?
– Nate
Commented Jul 14 at 7:02
• The original question is like this: Commented Jul 14 at 15:19
• There are 15 different medals in a drawer. (a) Suppose the medals in the drawer are divided into groups of 4, 5 and 6 medals. (i) Find the number of ways of dividing the medals into groups. (ii) If the three groups of medals are put into three boxes of different sizes, find the number of ways of distributing the medals. (b) Suppose the medals in the drawer are put into three boxes that can hold at most 3, 5 and 8 medals respectively. Find the number of ways of distributing the medals. Commented Jul 14 at 15:20

Another way to think about the problem (which simplifies the calculation) is to consider adding a 16th medal to the drawer and distributing all 16 across the 3 boxes.

The 16th medal acts as the 'missing medal' in one of the 3 boxes.

And hence the answer is (just as you calculated): $$C_{3,5,8}^{16} = \dfrac{16!}{3!5!8!} = 720720$$

I have no idea how they got $$3243240$$ but a couple ways of getting this number are: $$C_2^{15} \cdot C_5^{13} \cdot C_8^{9} + C_3^{15} \cdot C_4^{12} \cdot C_8^{9}$$

Or

$$C_3^{15} \cdot C_5^{12} \cdot C_8^{9}$$

• +1 : "Another way to think about the problem..." - nice. Commented Jul 14 at 7:42
• I have never seen combination written in such format! (3 items at the bottom) Is there any material I can read about it please? Commented Jul 14 at 15:21
• Honestly, I made it up because it felt intuitive and I haven't found any other uses of it online. Probably because (n,r) notation is generally preferred to nCr. However (n,r) notation is commonly extended to multinomial coefficients in much the same way en.wikipedia.org/wiki/Multinomial_theorem math.stackexchange.com/questions/4282492/… Commented Jul 14 at 16:44
• A friendly reminder that if you feel like your question has been sufficiently answered, you can accept an answer. This awards a green tick to let other users know the answer has been accepted and awards reputation to the author which helps incentivize further contributions. Commented Jul 27 at 5:01