CMA Foundation · Fundamentals of Business Mathematics and Statistics · Permutation and Combinations
A cost accounting department of a firm has 8 CMA trainees. A team of 3 trainees is to be chosen for a stock-verification visit. In how many different ways can the team be selected?
The team can be chosen in 56 ways. Since the order of choosing trainees is irrelevant, the count is the combination 8C3, which equals 8×7×6 divided by 3×2×1, giving 56. The figure 336 would arise only if order mattered.
- A24
- B56Correct
- C112
- D336
Explanation
Order does not matter, so use combinations: 8C3 = (8×7×6)/(3×2×1) = 56. The value 336 is 8P3, which wrongly counts the order of selection.
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