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 -
`(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`