tania brings 8.32 to the candy store. she wants to buy 21 jawbreakers does she have enough money? round your answer down to the
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This is a problem of Permutations. We have 3 cases depending on the number of B's. Since no more than three B's can be used we can use either one, two or three B's at a time.
Case 1: Five A's and One B
Total number of letters = 6
Total number of words possible =
Case 2: Five A's and Two B's
Total number of letters = 7
Total number of words possible =
Case 3: Five A's and Three B's
Total number of letters = 8
Total number of words possible =
Total number of possible words will be the sum of all three cases.
Therefore, the total number of words that can be written using exactly five A's and no more than three B's (and no other letters) are 6 + 21 + 56 = 83