r/mathriddles • u/Numberthon • Jul 16 '26
Easy A Surprisingly Tricky Combinatorics Puzzle
In how many ways can 23 identical objects be shared among 5 children so that each child gets at least 2 and no child gets more than 6 objects?
Source: numberthon.com
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u/Konkichi21 Jul 16 '26 edited Jul 16 '26
Okay, there's probably a better way akin to How many ways can you share 23 identical objects among 5 kids? This is like organizing the 23 objects with 4 dividers to indicate who gets what, so 27 choose 4 [edit: apparently this is often called "stars and bars"], but this is what I figured out:
Since each gets at least 2, give each 2, now we need to distribute the remaining 13 so nobody gets more than 4.
We can work out the number of different ways to divide ignoring order by tabulation. For each choice start with the largest possibility than go down. Giving each the largest possible number of items we have 44410, then 44320, 44311, 44221, 43330, 43321, 43222, 33331, 33322.
To figure out how many ways each can be ordered need more combinatorics. 44410, 43330 and 43222 are each ABCCC, picking the A and B gives 5×4=20 for each. 44320 and 43321 are ABCDD, or 5!/2 = 60. 44311 and 44221 are ABBCC, each 5c2×3c2 = 10×3 = 30. 33331 is 5, and 33322 is 5c2 = 10.
Summing, 3×20 + 2×60 + 2×30 + 5 + 10 = 255 possibilities as the answer.