MDM4U Permutations with some identical items 2016-02-11


Today we formalized counting permutations when there are some identical (indistinguishable) items.

For example, we discussed arranging the letters of the word BALL. If the two Ls were different in some way, there would be 4!=24 different arrangements of the letters. However, the Ls are not different, so we have counted every arrangement twice, for the two arrangements of the Ls (instead of only once). We can divide the number of arrangements in half to get the correct value, \frac{4!}{2}=12.

In general, if there are n items, with a group of p identical items, another group of q identical items, and so on, the number of different permutations of ALL n items is

\frac{n!}{p!q!\dots}

I assigned page 245 #1-6 from your textbook.

Want another explanation of the concept? Here’s a video to help. Mr. Richards uses different letters (a, b, c, instead of p, q), but it’s the same thing.

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