1. ## Probability help!

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.

c) The committee 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.

Case 1:

2 pizzas with 1 topping
4 pizzas with 2 toppings

Case 2:

3 pizzas with 1 topping
2 pizzas with 2 toppings
1 pizza with 3 toppings

Case 3:

4 pizzas with 1 topping
2 pizzas with 3 toppings

d) For one of these cases, determine the number of ways of choosing and distributing the 10 toppings.

I was lucky and found the exact same question on this page but I'm not sure I understand the answer.

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
I understand the "(10 choose 2)(8 choose 2)(6 choose 2)(4 choose 2)" but i'm not sure why we have to divide by 4! at the end.

It'd be great if someone could clarify this for me.

Thanks!!

2. bump can anyone help?

3. Originally Posted by ilc
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.

c) The committee 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.

Case 1:

2 pizzas with 1 topping
4 pizzas with 2 toppings

Case 2:

3 pizzas with 1 topping
2 pizzas with 2 toppings
1 pizza with 3 toppings

Case 3:

4 pizzas with 1 topping
2 pizzas with 3 toppings

d) For one of these cases, determine the number of ways of choosing and distributing the 10 toppings.

I was lucky and found the exact same question on this page but I'm not sure I understand the answer.

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
I understand the "(10 choose 2)(8 choose 2)(6 choose 2)(4 choose 2)" but i'm not sure why we have to divide by 4! at the end.

It'd be great if someone could clarify this for me.

Thanks!!
The counting scheme for case #1 takes the two pizzas with one topping each as a single group of 2 at the beginning, then each of the other 4 pizzas can be in any order, [notice the (2 choose 2) was deliberately omitted at the end, due to it's value being 1].

The counting scheme basically arranges the 4 pizzas with 2 toppings (inadvertently).
This gives 4! times the desired result.