<p>How would I solve a question such as: There are 25 people in a club, how many different combinations of 4 officers can be made?</p>
<p>25C4 = 25!/((25-4)!(4!)) = 25!/(21!4!) = (25 x 24 x 23 x 22)/(4 x 3 x 2 x 1) = 12650
The easiest way is to use your calculator:
Enter 25 on the main screen
Click “Math”
Go to “PRB”
Enter “nCr”
Enter 4 on the main screen
Click “Enter”</p>
<p>It is 25C4 (on the calculator, “25 nCr 4”) and not 25P4 (on the calculator, “25 nPr 4”) because combinations don’t take into account the order. Permutations do. This means that if Mary, John, Paul, and Jennifer are officials, that is one combination, period, not more than one like it would be if they were rearranged several times in different orders. “25 nPr 4” would yield a much bigger answer.</p>
<p>If worse comes to worst, you can just list the combinations out and count them. Usually the numbers won’t be big and you’ll be able to figure the question out by hand. The test-makers don’t expect you to know the nCr or nPr function.</p>