Comb = unordered where order doesen't matter.
perm = order matters.
But, look at this questions:
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?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 know I am wrong. Can someone clarify how I can make a better decision distinguishing the two.
Look at another one:
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?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?
Am I missing some key rule here? or am I just losing it!
Thanks,