Question

A fast-food restaurant offers 6 different burgers, 6 different side orders, 7 different flavor drinks, and 9 different flavors of ice cream. In how many ways can a combo containing 2 burgers, 3 different sides, 2 different flavor drinks, and 3 ice cream flavors be made?

Answer 1

Answer:

Total no of ways can a combo containing 2 burgers, 3 different sides, 2 different flavor drinks, and 3 ice cream flavors be made = 529,200

Step-by-step explanation:

Given -

A fast-food restaurant offers 6 different burgers, 6 different side orders, 7 different flavor drinks, and 9 different flavors of ice cream .

choose 2 burgers from 6 different burger = $\binom{6}{2}$

choose 3 different sides from  6 different side orders = $\binom{6}{3}$

choose 2 flavor drinks from 7 different flavor drinks = $\binom{7}{2}$

choose 3 ice cream from   9 different flavors of ice cream = $\binom{9}{3}$

Total no of ways can a combo containing 2 burgers, 3 different sides, 2 different flavor drinks, and 3 ice cream flavors be made=

= $\binom{6}{2}\times\binom{6}{3}\times\binom{7}{2}\times\binom{9}{3}$          ( Using combination )

= $\frac{6!}{(2!)(4!)}\times\frac{6!}{(3!)(3!)}\times\frac{7!}{(2!)(5!)}\times\frac{9!}{(3!)(6!)}$

= $15\times20\times21\times84$

= 529,200