CMA Foundation · Fundamentals of Business Mathematics and Statistics · Permutation and Combinations
A firm has 6 chartered accountants and 4 cost accountants. A team of 4 members must be chosen with exactly 2 from each group. How many teams are possible?
There are 90 possible teams. Choosing 2 of the 6 chartered accountants gives 6C2 = 15 ways, and choosing 2 of the 4 cost accountants gives 4C2 = 6 ways. Multiplying gives 15×6 = 90. Using 10C4 = 210 would ignore the required split.
- A90Correct
- B210
- C60
- D15
Explanation
Choose 2 of 6: 6C2 = 15. Choose 2 of 4: 4C2 = 6. Teams = 15×6 = 90. The value 210 is 10C4, which ignores the 2-and-2 requirement.
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