# Unit Key Question - Probability Pizza :)

• Jan 9th 2009, 06:13 PM
cnmath16
Unit Key Question - Probability Pizza :)
The student council is ordering pizza for their next meeting. There are 20 council members, 7 of whom are vegetarian. A committee of 3 will order 6 pizzas from a pizza shop that has a special price for large pizzas with up to three toppings. The shop offers 10 different toppings.

a) How many different pizza committees can the council choose if there must be at least 1 vegetarian and 1 non-vegetarian on the committee?

b) In how many ways could the committee choose up to 3 toppings for a pizza?

c) The commitee wants as much variety as possible in the toppings. They decide to order each topping exactly once and to have at least 1 topping on each pizza. Describe the different cases possible when distributing the toppings in this way.

any help for any of these questions would be great.
THANKS!
• Jan 9th 2009, 11:45 PM
Combinatorics
Hello cnmath16
Quote:

Originally Posted by cnmath16
The student council is ordering pizza for their next meeting. There are 20 council members, 7 of whom are vegetarian. A committee of 3 will order 6 pizzas from a pizza shop that has a special price for large pizzas with up to three toppings. The shop offers 10 different toppings.

a) How many different pizza committees can the council choose if there must be at least 1 vegetarian and 1 non-vegetarian on the committee?

b) In how many ways could the committee choose up to 3 toppings for a pizza?

c) The commitee wants as much variety as possible in the toppings. They decide to order each topping exactly once and to have at least 1 topping on each pizza. Describe the different cases possible when distributing the toppings in this way.

any help for any of these questions would be great.
THANKS!

(a) There are 13 ways of choosing the non-veg member and 7 ways of choosing the veg. Once these are chosen, there are then 18 ways of choosing the third member. Total number of ways so far = 13 x 7 x 18. But in each of these combinations, the same group of three will appear twice: once when a particular member was chosen specifically to represent their group (v or non-v) and once when they were chosen freely as the third person. So you must divide the total number of ways by 2.

(b) Add together the number of ways of choosing exactly one topping, exactly two toppings and exactly three toppings. So that's choosing 1 item from 10, 2 items from 10, 3 items from 10. OK?

(c) There are 10 toppings and 6 pizzas. Each pizza must have at least one, and no more than 3 toppings, and all 10 toppings are used exactly once. So the following combinations are possible:

• 2 x 1 and 4 x 2 (i.e. 2 pizzas with 1 topping each, and 4 pizzas with 2 toppings each)
• 3 x 1, 2 x 2 and 1 x 3
• 4 x 1, 2 x 3

• Jan 12th 2009, 03:30 PM
cnmath16
Thanks for much for your reply. Part d of this question states, "For one of these cases, determine the number of ways of choosing and distributing the 10 toppings"

I was wondering if you would be able to check my answer for me.

For Case #1: (2 pizzas with 1 topping each and 4 pizzas with exactly 2 toppings on each)

(10 choose 2)(8 choose 2)(6 choose 2)(4 choose 2) ÷ 4!

= 45 x 28 x 25 x 6 ÷ 24
= 4275

Therefore, For Case #1..there are 4275 ways of choosing and distributing the ten toppings.

THANKS
• Jan 12th 2009, 10:40 PM