+0  
 
+3
2329
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avatar+210 

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.

 Aug 14, 2018
 #1
avatar+210 
+3

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.

 Aug 14, 2018
 #2
avatar+129852 
+6

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

 Aug 14, 2018

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