GRE Prep Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Mario has a *fabulous* drug collection. Tomorrow he plans
[#permalink]
26 Aug 2015, 08:06
26 Aug 2015, 08:06
1
1
4
Bookmarks
00:00
Question Stats:
45% (02:18) correct
54% (02:44) wrong based on 24 sessions
Hide Show timer Statistics
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
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
Show: :: My reasoning
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 \(\frac{4!}{(2!*2!)}*\frac{8!}{(2!*6!)}=168\).
Show: :: Supplied answer
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.
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: Combination Problem Struggles
[#permalink]
10 Feb 2016, 09:30
10 Feb 2016, 09:30
please post answer for this and explain the concept behind
Re: Mario has a *fabulous* drug collection. Tomorrow he plans
[#permalink]
03 Jan 2018, 00:09
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
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
Show: :: My reasoning
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 \(\frac{4!}{(2!*2!)}*\frac{8!}{(2!*6!)}=168\).
Show: :: Supplied answer
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.
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.
Can anyone please explain this>?
