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Undo revision 417372 by Special:Contributions/HelpJazz (User talk:HelpJazz) let's see if I can't get this to work
For example, the letters ''p'' and ''e'' in the word "pepper" are indistinguishable from one another. There are 6 total letters, which are comprised of 3 <i>p</i>'s, 2 <i>e</i>'s, and 1 ''r''. The total number of permutations would thus be 6! / (3 ! 2! 1!), or 60.
For example, the letters ''p'' and ''e'' in the word "pepper" are indistinguishable from one another. There are 6 total letters, which are comprised of 3 <i>p</i>'s, 2 <i>e</i>'s, and 1 ''r''. The total number of permutations would thus be 6! / (3 ! 2! 1!), or 60.
===Combinations===
It is often useful to find out how many groups of ''r'' objects can be created out of a group of ''n'' objects. This can be calculated from a function called "choose" (read aloud as "n choose r"):
:<math>\begin{pmatrix} n \\ r \end{pmatrix} = \frac{n!}{(n-r)!r!}</math>
For example, take a race which has 10 competitors. There are 10 choose 3 different combinations of medal winners (that is, racers who get either 1st, 2nd, or 3rd place), which is a total of 10! / (10 - 3)! • 3! = 10! / 7! • 3! = 120 different combinations.
[[category: mathematics]]
[[category: mathematics]]