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In how many ways can at least (4) and at most (6) gifts be selected from (14) different gifts?

Answer and explanation

Correct answer: (6006)

The direct answer is option C, 6006. “At least 4 and at most 6” permits selecting exactly 4, 5, or 6 gifts. Because order does not matter, add the three combination counts: \(\binom{14}{4}+\binom{14}{5}+\binom{14}{6}=1001+2002+3003=6006\). Option A, 4004, omits part of the permitted selections. Option B, 5005, is also too small and does not equal the full sum. Option C includes every allowed size once and is correct. Option D, 7007, overcounts. Do not use permutations because arranging selected gifts is irrelevant; only the chosen set matters. Exam cue: translate the interval directly into a sum from the lower number through the upper number.

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CombinationsClass11Hard

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