Giving twelve pencils so everyone gets one

Question

You distribute 12 identical pencils to 4 students, and each student must receive at least 1 pencil. In how many ways can the pencils be distributed?

Accepted Answer

The "at least one each" condition is handled by first giving every student one pencil up front. That uses 4 pencils and guarantees the minimum. Now 12 - 4 = 8 pencils remain to distribute freely, with no restriction (a student may get none of the remaining ones). So we need non-negative counts summing to 8 across 4 students, which by stars and bars is \binom{8 + 4 - 1}{4 - 1} = \binom{11}{3} = \frac{11 	imes 10 	imes 9}{6} = 165. Equivalently, this is \binom{n-1}{k-1} = \binom{11}{3} for the "each at least one" case with n = 12, k = 4. Takeaway. For "each group at least one," hand out one p…

Giving twelve pencils so everyone gets one

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