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January 30, 2015

January 30, 2015

Posted by **Lisa** on Saturday, January 19, 2008 at 10:59pm.

a) 7-letter words

b) 7-letter word beggining with MA

c) 4-letter words

i) all letters different

ii) 2 letters same,2 different

iii) 2 pairs same letters

iv) 3 letters same

- data management -
**Damon**, Sunday, January 20, 2008 at 1:15pmI am not doing the problem for you but perhaps I can give you an idea.

Let's see. I have to work with

M M M A A L S

If all letters were different the answer would be 7!, but they are not.

Let me try a simpler example

Say

A A B B

A B A B

A B B A

B A A B

B B A A

B A B A Hmm six of them, not 4!

Looks like permutations of 4 items divided by permutations of 2 As divided by permutations of 2 Bs

4!/(2!*2!) = 4*3*2 /( 2*2) = 6

Let's try it with 3 As and 1 B

A A A B

A A B A

A B A A

B A A A Is that 4!/(3!*1!) ?

4*3*2 /(3*2*1) = 4, sure enough.

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