CMA Foundation · Fundamentals of Business Mathematics and Statistics · Permutation and Combinations
From 6 men and 4 women, a panel of 5 persons is to be chosen so that it contains at least 2 women. How many panels are possible?
There are 186 panels. Total unrestricted panels are 10C5 = 252. Remove those with no woman (6C5 = 6) and those with exactly one woman (4×6C4 = 60). Thus 252 − 66 = 186 panels contain at least two women.
- A186Correct
- B210
- C120
- D252
Explanation
Total panels = 10C5 = 252. Subtract panels with 0 women = 6C5 = 6, and with 1 woman = 4C1×6C4 = 4×15 = 60. So 252 − 6 − 60 = 186. The value 210 results from forgetting to remove the 0-woman case... actually 252−42 is not used; 210 is a plain distractor, and 120 counts only 2W+3M-type cases incorrectly.
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