How many ways are there to choose 3 scoops of ice cream from 8 available flavours, if: You can choose the same flavour more than once, and The order of scoops does not matter (i.e., chocolate, vanilla, chocolate is the same as chocolate, chocolate, vanilla)?

Question

BlackTom AI · Accepted Answer

We need to count the number of ways to choose 3 scoops from 8 flavours when repetitions are allowed and order doesn’t matter. This is a classic stars-and-bars (multiset) problem. The formula for combinations with repetition is C(n + k − 1, k), where n is the number of flavours and k is the number of items chosen. Here, n = 8 and k = 3, so the count is C(8 + 3 − 1, 3) = C(10, 3) = 120.

Now evaluate each option:

- Option a: says 11^3. This treats each scoop as an independent choice among 11 possibilities and preserves order, which is not our scenario. Our problem restricts to 3 scoops total and ignores order, plus repetitions are allowed in a specific constrained way; exponential counting is inappropriate here.

- Option b: says 10^3. Similar issue as option a: it implies ordered triples with replacement from 10 options, which overcounts and also ignores the indistinguishability of the scoops in the final arrangement.

- Option c: uses binomial coefficient binom{11}{3}. This would correspond to a different counting model (e.g., choosing 3 items from 11 without repetition) and does not reflect the repetition allowance in a 3-choice multiset from 8 flavours.

- Option d: uses binom{10}{3} with a prefactor (103) in front, matching the standard result binom{8+3−1}{3} = binom{10}{3}. The structure aligns with the stars-and-bars approach for multisets of size 3 from 8 types.

- Option e: states 8^3. This would count ordered triples of scoops with 8 options each, which overcounts because order is irrelevant and we’re not distinguishing permutations of the same multiset.

- Option f: expresses binom{8}{3} with an extra (83) factor, which corresponds to choosing 3 distinct flavours without repetition, ignoring the possibility of repeating a flavour. That’s not appropriate here since repeating flavours is allowed (e.g., two chocolate and one vanilla).

In summary, the correct counting method for this problem is the combinations with repetition formula, yielding binom{10}{3}. The other options either assume order, misuse the repetition allowance, or misinterpret the counting model (with or without repetition) for this scenario.

How many ways are there to choose 3 scoops of ice cream from 8 available flavours, if: You can choose the same flavour more than once, and The order of scoops does not matter (i.e., chocolate, vanilla, chocolate is the same as chocolate, chocolate, vanilla)?单项选择题

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