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From (8) friends (4) are to be selected for a trip and at least one of two friends must go. How many ways are there?

Answer and explanation

Correct answer: (55)

The direct answer is option B, 55. We select 4 friends from 8, and at least one of two particular friends must be selected. Use the complement. First count all four-friend groups: C4 = 8!/(4!4!) = 70. Now count the groups in which neither of the two particular friends goes. Then all 4 members must come from the other 6 friends, giving C4 = 6!/(4!2!) = 15. Subtracting gives 70 − 15 = 55. Option A, 45, subtracts an incorrect number. Option B, 55, is correct because it excludes exactly the groups missing both required candidates. Option C, 60, does not remove all such groups. Option D, 70, is the unrestricted total and includes groups where both particular friends are absent. “At least one” allows exactly one or both, so we must not exclude groups containing both. Memory cue: at least one of a pair = total groups minus groups containing neither.

Related tags

CombinationsClass11Medium

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