Let's do 5 letter words first. You're concerned about 2 groups of letters; those groups without duplicates and those with duplicates (the "a"). There is only one group of 5 letters without duplicates. You can arrange those 5! different ways or 120.
Now let's consider the 5 letter groups with 2 a's. You need 3 more letters from the remaining four, i.e. (4 choose 3) which is 4. You can arrange those groups of 5 letters in 5! /2! different ways (60).
So the number of 5 letter words is 120 + (4 * 60) = 360.
4 letter words:
Number of words without duplicates: Choose 4 letters from the possible 5 unique letters (5 different ways) and multiple by the number of ways you can arrange them (4!). That's 5*24 = 120.
Number of words with duplicate a's: Use 2 a's plus 2 of the remaining 4 (4 choose 2). You can arrange those groups 4!/2! ways. That's 6*12 = 72. Add that to the 120 and you have 192 four letter words.
3 letter words:
Number of words without duplicates: Choose 3 letters from the possible 5 unique letters (10) and multiple by the number of ways you can arrange them (3!). That's 10*6=60
Number of words with duplicate a's: Use 2 a's plus 1 of the remaining 4 (4 choose 1). You can arrange those groups 3!/2! ways. That's 4*3 = 12. Add that to the 60 and you have 72 three letter words.
2 letter words:
Number of words without duplicates: Choose 2 letters from the possible 5 unique letters (10) and multiple by the number of ways you can arrange them (2!). That's 10*2=20
Number of words with duplicate a's: Just 1 (aa). Add that to the 20 and you have 21 two letter words.
Add them all together and you have your answer.
]]>But be careful of duplicate words, because there are two "a"s!
]]>I know how to get all the 6 letter words, 6!/2!, and the 1 letter words, obviously 5, but is there an easier way to get all the 2,3,4 and 5 letter words instead of just counting?
]]>