Question

a company sells regular hamburgers as well as a larger burger. either type can include cheese, relish, lettuce, tomato, mustard, or ketchup.
a. how many different hamburgers can be ordered with exactly 5 extras?
b. how many different regular hamburgers can be made that use any 5 of the extras?
c. how many different regular hamburgers can be ordered with at least 3 extras?

a. there are □ different hamburgers that can be made using exactly 5 of the extras.

Explanation:

Step1: Choose extras (combination)

First, calculate the number of ways to choose 5 extras from 6. The combination formula is $C(n,k)=\frac{n!}{k!(n-k)!}$, so $C(6,5)=\frac{6!}{5!(6-5)!}=6$.

Step2: Choose burger type

There are 2 burger types (regular, larger). Multiply the number of extra combinations by the number of burger types: $6\times2=12$.

Step3: Regular burger, choose 5 extras

For regular burgers only, use the combination result directly: $C(6,5)=6$.

Step4: Calculate at least 3 extras

"At least 3" means 3, 4, 5, or 6 extras. Calculate each combination:
$C(6,3)=\frac{6!}{3!(6-3)!}=20$, $C(6,4)=\frac{6!}{4!(6-4)!}=15$, $C(6,5)=6$, $C(6,6)=\frac{6!}{6!(6-6)!}=1$.
Sum these values: $20+15+6+1=42$.

Answer:

a. 12
b. 6
c. 42