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From the numbers 1 to 25, how many ways are there to choose 3 odd and 2 even numbers?

Answer and explanation

Correct answer: 18876

Use separate combinations for the two parity groups. From 1 through 25 there are 13 odd numbers and 12 even numbers. The number of ways to choose 3 odd numbers is C(13,3) = 13 × 12 × 11 ÷ 6 = 286. The number of ways to choose 2 even numbers is C(12,2) = 12 × 11 ÷ 2 = 66. Since one selection from each group is required, multiply the independent counts: 286 × 66 = 18,876. Thus option C is correct. The alternatives are not equal to this product; they can result from omitting a parity condition, using an incorrect group size, or applying a non-combination count.

Related tags

CombinationsOdd Even NumbersRestricted SelectionPermutations And CombinationsMathematicsClass 11 Mcq

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