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

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

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"


cool cool cool

CPhill  Aug 14, 2018

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