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In how many ways can at least (4) and at most (6) gifts be selected from (15) different gifts?

Answer and explanation

Correct answer: (9373)

The direct answer is option C, 9373. The permitted choices are exactly 4, 5, or 6 gifts. Therefore calculate \(\binom{15}{4}+\binom{15}{5}+\binom{15}{6}\). The values are \(1365+3003+5005=9373\). Option A, 8008, is smaller than the complete sum and misses valid selections. Option B, 9009, is also not the sum. Option C is the exact total. Option D, 12376, overcounts and includes choices outside the stated range or uses an incorrect calculation. Since gifts are different but selection order is irrelevant, combinations—not permutations—are required. A useful check is that the answer must be greater than the largest single term 5005 but less than the sum of many broader choices.

Related tags

CombinationsClass11Hard

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