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Joe wants to find all the four-letter words that begin and end with the same letter. How many combinations of letters satisfy this property?

BTW If you're reading this you're a legend.

ColdplayMX Aug 14, 2018

#1**+2 **

Is the answer 8650, because there are 26 letters to choose from to use as starting letters, and then 26 choose 2 ways to determine the middle two letters, giving us C(26, 2) * 26 = 8650????

I'm not sure.

ColdplayMX Aug 14, 2018

#2**+2 **

We have 26 ways to choose the first and last letters

Coldplay. you don't specify if the intervening letters can be repeats of the first [and last] chosen letter....but...if we don't have any restrictions on the intervening letters.....we have 26 choices for the first letter, 26 for the second, 26 for the third...and only 1 for the last [ whichever letter you chose first ] = 26^3 = 17576 "words"

CPhill Aug 14, 2018