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April 20, 2014

April 20, 2014

Posted by **avm** on Wednesday, September 23, 2009 at 8:56pm.

- Gr. 12 Data Management -
**MathMate**, Wednesday, September 23, 2009 at 10:48pmThe number of arrangements for n distinct lettes is

n!

For example: 4!=24 distinct words can be made from the letters ABCD.

The number of arrangements for n letters, of which p are identical is

n!/p!

For example: 4!/2!=12 words can be made from the letters AABC.

The number of arrangements for n letters, of which p are identical and q are identical is

n!/(p!q!)

For example: 6!/(2!2!)=180 words can be made from the letters AABBCD.

If two letters have to be together all the timee, treat them as a single letter.

For the word BASKETBALL,

first arrange them in alphabetical order:

AABBEKLLST

Out of the 10 letters, AA and BB are repetitions, LL can be treated as one single letter (to have a total of 11 letters).

So the number of distinct words possible is

11!/(...)

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