How many "words" can be written using exactly five A's and no more than three B's (and no other letters)?

asked Aug 1, 2019 162k views

4 votes

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5 votes

5 A's and no B's -

5 A's and 1 B -
(6!)/(5!)=6

5 A's and 2 B's -
$(7!)/(5!2!)=(6\cdot7)/(2)=21$

5 A's and 3 B's -
$(8!)/(5!3!)=(6\cdot7\cdot8)/(2\cdot3)=56$

56+21+6+1=84

Venomoustoad answered Aug 6, 2019

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