Respuesta :
Answer:
a) 40 different hamburgers can be ordered with exactly three extras
b) 20 different regular hamburgers can be ordered with exactly three extras
c) 7 different regular hamburgers can be ordered with at least five extras
Step-by-step explanation:
The order in which the extras are ordered is not important. So we use the combinations formula to solve this question.
Combinations formula:
[tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
In this problem:
2 options of hamburger(regular or larger)
6 options of extras(cheese, relish, lettuce, tomato, mustard, or catsup.).
(a) How many different hamburgers can be ordered with exactly three extras?
1 hamburger type, from a set of 2.
3 extras, from a set of 6. So
[tex]C_{2,1}*C_{6,3} = \frac{2!}{1!(2-1)!}*\frac{6!}{3!(6-3)!} = 2*20 = 40[/tex]
40 different hamburgers can be ordered with exactly three extras
(b) How many different regular hamburgers can be ordered with exactly three extras?
3 extras, from a set of 6. So
[tex]C_{6,3} = \frac{6!}{3!(6-3)!} = 20[/tex]
20 different regular hamburgers can be ordered with exactly three extras
(c) How many different regular hamburgers can be ordered with at least five extras?
Five extras:
5 extras, from a set of 6. So
[tex]C_{6,5} = \frac{6!}{5!(6-5)!} = 6[/tex]
Six extras:
6 extras, from a set of 6. So
[tex]C_{6,6} = \frac{6!}{6!(6-6)!} = 1[/tex]
6 + 1 = 7
7 different regular hamburgers can be ordered with at least five extras
