dance4ever1 dance4ever1
  • 05-08-2018
  • Mathematics
contestada

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

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konrad509
konrad509 konrad509
  • 05-08-2018

5 A's and no B's - [tex] 1 [/tex]

5 A's and 1 B - [tex] \dfrac{6!}{5!}=6 [/tex]

5 A's and 2 B's - [tex] \dfrac{7!}{5!2!}=\dfrac{6\cdot7}{2}=21 [/tex]

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

[tex]56+21+6+1=84 [/tex]

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