Basic statistics?

Question

Of the 6 courses offered by the music department at her college, Kay must choose exactly 4 of them. How many different combinations of 4 courses are possible for Kay if there are no restrictions on which 4 courses she can choose?

Please explain step by step. I'm just setting foot into this area of mathematics. I understand that the number of possible permutations are

6*5*4*3 = 360.

I want to know the way to find the different combinations including repeats.

Like...

A B C D = 1
A B D C = 2
.... = 3


... = blah?

What's the mathematical way to find the number of those?

If there were two choices the number of combinations would be

A B = 1
B A = 2

Answer 1

Yes, you got the permutations right nPr= 360
Now for combinations, or nCr you are trying to figure out six things taken four at a time
This is the formula
nCr=n!/r!(n-r)!
n! represents n factorial which means the following
n(n-1)(n-2)(n-3)(n-4)...........3,2,1
Therefore, for your problem, you have 6C4
6! / 4!(6-4)!
6*5*4*3*2*1 / 4*3*2*1*2*1
Before you get hasty and start multiplying everything out, look for shortcuts. You can actually cancel out the 4*3*2 in the numerator and denominator and just eliminate the 1's because they don't matter.
6*5/2
30/2=15
There are 15 possible combinations.

Answer 2

since there are 4 courses out of 6, the only combinations I could think of is:
6 x 4 = 24

why make it so hard? :(

Answer 3

This is a combination of 6 items taken 4 at a time.
The formula for the total would be 6!/(4!2!)=15.
As for the enumeration, a few are ABCD, ABCE, ABCF, ABDE, ABDF, ABEF, ACDE, ACDF, ACEF, ADEF, BCDE, BCDF, BDEF, CDEF

Answer 4

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