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# Permutations & Combinations

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Suppose letters of the alphabet are chosen at random to make up words. How many four letter words are possible, if the first and last letter are the same?

Also, how did you know what equation to use? What's an easy way to remember which to use when?

Feb 16, 2020

#1
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Now, I'm assuming that any 4-letter "word" is valid such as: ACST. Ordinarily, if that is the case, then you would have: 26P4 =26 x 25 x 24 x 23 =358,800 "words".

But, because you want the first and last letters the same, then will use an example to illustrate what you want. Take the word "ROAR". You have 26 choices for the first "R", 25 choices for "O", 24 choices for "A", and because we are repeating the first "R" again, you have 26 choices.

So: 26 x 25 x 24 x 26 =405,600 "words"

Feb 16, 2020
#2
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Your answer is not correct because 26P4 does not allow for repeats.

It is basically saying you have 26 letters and you must choose 4 of them (where the order is important)

I know you have gone on to modify it but you have still kept the same error.

Suppose letters of the alphabet are chosen at random to make up words. How many four letter words are possible, if the first and last letter are the same?

Also, how did you know what equation to use? What's an easy way to remember which to use when?

I'll give it a go.

26 choices for the first letter, 26choices for the second, 26 choices for the third and only 1 choice at the end.

26^3 = 17576 possible 'words'

Feb 16, 2020