Rigshim
A person has 9 friends, 4 of whom are male. In how many ways can he plan a trip so that there are at least 2 other male friends, but not more than 3 females?
(A) 32 (B) 96 (C) 286 (D) 120 (E) 250
Hello,
Rigshim. You might want to post this question in the PS Quant forum so that it has a timer affixed to it and members can try it out. For my part, I calculated (C), 286, as the answer, tracing the logic below:
1) There can be 2, 3, or 4 males within the given constraints.
2) There can be 3, 2, 1,
or 0 females within the given constraints.
3) Run the combinations for males and females independently: 4C2, 4C3, 4C4 and 5C3, 5C2, 5C1, 5C0. This yields 6, 4, 1, 10, 10, 5, and 1, respectively. (You have to remember that 0! = 1.)
4) Multiply each "male" count with each "female" count and add:
\(6*10+6*10+6*5+6*1=156\)
\(4*10+4*10+4*5+4*1=104\)
\(1*10+1*10+1*5+1*1=26\)
\(156+104+26=286\)
This was just the first way my mind thought to work the problem. I am not sure what the source may be, but to better conform to what might appear on the GMAT™, I would suggest reordering the answers to count up (i.e. shift (D) and (E) one place to the left or up and make (C) (E) instead). Anyway, thank you for sharing, and good luck with your studies.
- Andrew