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 4 extras? b. How many different regular hamburgers can be made that use any 4 of the extras? c. How many different regular hamburgers can be ordered with at least 3 extras?

a. 30

b. 15

c. 42

Step-by-step explanation:

We have two types of hamburgers and six types of extras.

a.

To find the number of different burgers with 4 extras, we have a combination of 6 choose 4 to find the number of possibilities for the extras, and we have that amount of possibilities for each of the two types of burger, so we have:

2 * C(6,4) = 2 * 6! / (4! * 2!) = 2 * 6 * 5 / 2 = 30 different hamburgers.

b.

The number of different regular hamburgers is half the number of different burgers found above:

C(6,4) = 6! / (4! * 2!) = 6 * 5 / 2 = 15 different hamburgers.

c.

We want at least 3 extras, so we can have 3, 4, 5 or 6 extras, then we need to calculate combinations of 6 choose 3, 4, 5 and 6:

3 extras: C(6,3) = 6! / (3! * 3!) = 6 * 5 * 4 / 3 * 2 = 20 different hamburgers.

4 extras: C(6,4) = 6! / (4! * 2!) = 6 * 5 / 2 = 15 different hamburgers.

5 extras: C(6,5) = 6! / (5! * 1!) = 6 / 1 = 6 different hamburgers.

6 extras: C(6,6) = 6! / (6! * 0!) = 1 different hamburger.

Total: 20 + 15 + 6 + 1 = 42 different regular burgers

=0

Step-by-step explanation: