Question

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

Answer 1

5 A's and no B's - $1$

5 A's and 1 B - $\dfrac{6!}{5!}=6$

5 A's and 2 B's - $\dfrac{7!}{5!2!}=\dfrac{6\cdot7}{2}=21$

5 A's and 3 B's - $\dfrac{8!}{5!3!}=\dfrac{6\cdot7\cdot8}{2\cdot3}=56$

$56+21+6+1=84$