Question

A fast-food restaurant offers 6 different burgers, 5 different side orders, 8 different flavor drinks, and 8 different flavors of ice cream. In how many ways can a combo containing 2 burgers, 2 different sides, 3 different flavor drinks, and 2 ice cream flavors be made?

Answer 1

Answer:

235,200 ways

Step-by-step explanation:

A fast-food restaurant offers 6 different burgers

5 different side orders

8 different flavor drinks

8 different flavors of ice cream.

use combination to find the ways nCr

$nCr= \frac{n!}{r!(n-r)!}$

2 burgers can be selected from 6 different burgers in 6C2 ways

$6C2= \frac{6!}{2!(6-2)!}=15$

2 different sides can be selected from 5 different side orders in 5C2 ways

$5C2= \frac{5!}{2!(5-2)!}=10$

3 different flavor drinks can be selected from 8 different flavor drinks in 8C3ways

$8C3= \frac{8!}{3!(8-3)!}=56$

2 ice cream flavors can be selected from 8 ice cream flavors in 8C2 ways

$8C2= \frac{8!}{2!(8-2)!}=28$

Total ways = $15 \cdot 10 \cdot 56 \cdot 28=235200$