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In how many ways can at least (3) and at most (4) gifts be selected from (12) different gifts?

Answer and explanation

Correct answer: 715

The selection may contain either 3 gifts or 4 gifts. Hence, the total number of ways is \(\binom{12}{3}+\binom{12}{4}=220+495=715\). Option 495 counts only the selections of 4 gifts, so it omits the selections of 3 gifts. Exam tip: “At least 3 and at most 4” means add the cases for 3 and 4 selections.

Related tags

CombinationsPermutations And CombinationsClass 11SelectionCounting PrinciplesBinomial CoefficientsMathematicsClass 11 Mcq

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