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A box has (4) red, (3) blue, and (2) green distinct balls. Three balls are drawn in order without replacement. The first ball must not be red and the third must be green. How many ways are possible?

Answer and explanation

Correct answer: (72)

Fix the third ball as green and count first-place validity separately. The middle choices differ when the first ball is green or blue.

Related tags

PermutationsCombinationsCountingHardFundamental Principle Of CountingPermutations And CombinationsMathematicsClass 11 Mcq

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