Here's a question I got an ETS practice exam that I couldn't figure out how they got their answer. I rewrote it and the wrong answers to keep the copyright police away.
Mario has a *fabulous* drug collection. Tomorrow he plans to go to a festival and get stoned on
four different drugs out of ten he selected. Mario wants to party all day so he needs to
take at least two stimulants from the four uppers out of the ten he brought.
How many different possible drug combinations are there for Mario to get as high as a Georgia pine?
A. 96
B. 108
C. 115
D. 156
E. 208
I figured the problem required the product of two instances of the combinations formula, one for the selection from the four stims and one for the remaining drugs. Using \(\frac{n!}{(k!*(n-k)!)}\) I got \(\frac{4!}{(2!*2!)}*\frac{8!}{(2!*6!)}=168\).
ETS gave the answer as C. 115
I can't figure out how this is possible since the prime factorization of 115 is 23 and 5, and the 23 should not be accessible through the factorial combination formula. They must have used some addition/subtraction to reach this number.