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Mario has a *fabulous* drug collection. Tomorrow he plans [#permalink]
26 Aug 2015, 08:06

1

This post received KUDOS

00:00

Question Stats:

0% (00:00) correct
100% (02:27) wrong based on 3 sessions

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.

Re: Mario has a *fabulous* drug collection. Tomorrow he plans [#permalink]
03 Jan 2018, 00:09

Anki wrote:

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.

Re: Mario has a *fabulous* drug collection. Tomorrow he plans [#permalink]
08 Jan 2018, 03:24

1

This post received KUDOS

nice choice of question!

George must take 4 drugs from a collection of 10 However, he must have at least 2 from the four uppers and some variation from the regular He can either have 2 upper or 3 upper or a 4 upper. In case he has 2 upper he will have to take 2 regular; If 3 upper than 1 regular or 4 upper and no regular This boils down to (6c2 * 4c1) + (6c1* 4c3) + 4c4 = 115
_________________

This is my response to the question and may be incorrect. Feel free to rectify any mistakes

greprepclubot

Re: Mario has a *fabulous* drug collection. Tomorrow he plans
[#permalink]
08 Jan 2018, 03:24