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| + | This is a great way to think about it, and it's how I did it. | ||
| + | --[[User:Norlow|Norlow]] 22:14, 17 September 2008 (UTC) | ||
Latest revision as of 18:14, 17 September 2008
How many permutations of the letters ABCDEFGH
A. Contain ED
- I thought of ED as one letter and the rest as separate.
then there is only one way to make ED = 1! and 7! ways to rearrange them together= 7!
Answer: 7!1!= 5040
I think the same concept can be used for the rest.
E .
- you can count CABED as one word
4!
F. Not possible 0
This is a great way to think about it, and it's how I did it. --Norlow 22:14, 17 September 2008 (UTC)
