Buying six doughnuts from four types

Question

A bakery offers 4 types of doughnut with plenty of each in stock. You buy a box of 6 doughnuts. Only how many of each type you take matters, not the order, and you may take several of the same type or none of some types. How many different boxes of 6 are possible?

Accepted Answer

A box is described by how many of each type you choose: counts (a, b, c, d) of non-negative whole numbers with a + b + c + d = 6. This is the classic "how many of each" selection, which stars and bars counts directly. Use 6 stars (doughnuts) and 3 bars (separating the 4 types). Arrange 6 + 3 = 9 symbols and choose the 3 bar positions: \binom{6 + 4 - 1}{4 - 1} = \binom{9}{3} = \frac{9 	imes 8 	imes 7}{6} = 84. Takeaway. Choosing n items from k types with repetition allowed (a multiset) is \binom{n+k-1}{k-1}. Six doughnuts from four types give \binom{9}{3} = 84.

Buying six doughnuts from four types

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