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distinguish comb/perm

Problem Solving — algebra and arithmetic (GMAT Focus Edition)
Expert replies
by Gurpinder » Tue Nov 08, 2011 2:08 pm
Hi guys, having a little trouble distinguishing the two.

Comb = unordered where order doesen't matter.
perm = order matters.

But, look at this questions:
Question wrote: A certain university will select 1 of 7 candidates eligible to fill a position in the mathematics department and 2 of 10 candidates eligible to fill 2 identical positions in the computer science department. If none of the candidates is eligible for a position in both departments, how many different sets of 3 candidates are there to fill the 3 positions?
To me, the second part of the question looks like a perm. Since there are 2 seats and therefore 2 candidates (hence they are distinguishable) wouldn't it be 10!/8!=90?

I know I am wrong. Can someone clarify how I can make a better decision distinguishing the two.

Look at another one:
The principal of a high school needs to schedule observations of 6 teachers. she plans to visit one teacher each day for a work week (M-F) so will only have time to see 5 of the teachers. How many different observation schedules can she create?
To me there is no order here as the principal can meet any of the 5 teachers any day! The order in which she meets those "5" selected teachers does not matter. Hence, shouldnt this be a combination problem?

Am I missing some key rule here? or am I just losing it!

Thanks,
"Do not confuse motion and progress. A rocking horse keeps moving but does not make any progress."
- Alfred A. Montapert, Philosopher.
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Source: — Quantitative Reasoning |

by shankar.ashwin » Tue Nov 08, 2011 9:16 pm
Read the question carefully, the second part states that the 2 students fill up identical positions.(there is not differentiation between both of them). Its similar to "In how many ways can you pick 2 people from a group of 10". Here obviously the order you pick them up does not matter.

Since, given its an identical position, this would be a combination problem.

7C1 * 10C2 = 315 IMO
Gurpinder wrote:
Question wrote: A certain university will select 1 of 7 candidates eligible to fill a position in the mathematics department and 2 of 10 candidates eligible to fill 2 identical positions in the computer science department. If none of the candidates is eligible for a position in both departments, how many different sets of 3 candidates are there to fill the 3 positions?
I think Pete clarified your other question..
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