# Combinations

• Dec 6th 2010, 06:30 PM
tmas
Combinations
I can't get the last question.....

A pizza company advertises that it has 15 toppings from which to choose (There are six vegetable toppings)

e) how many different 3 topping pizzas have at least one vegetable?
This is what i did:
6nCr3 = 20 15nCr3 = 455
455-19 = 436

The answer in the back of the book is:371
• Dec 6th 2010, 06:54 PM
Soroban
Hello, tmas!

Quote:

A pizza company advertises that it has 15 toppings from which to choose.
There are six vegetable toppings.

e) How many different three-topping pizzas have at least one vegetable?

There are 6 Vegetable toppings and 6 Others.

There are: . $_{15}C_3 \:=\:\dfrac{15!}{3!\,12!} \:=\:455$ possible three-topping pizzas.

The opposite of "at least one vegetable" is "no vegetables."

There are: . $_9C_3 \:=\:\dfrac{9!}{3!\,6!} \:=\:84$ three-topiing pizzas with no vegetables.

Therefore, there are: . $4565 - 84 \:=\:371$ pizzas with at least one vegetable.

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

If you want to list the pizzas with at least one vegetable,
. . be sure you make a complete list.

One vegetable and two Others
, , $(_6C_1)(_9C_2) \:=\:(6)(36) \:=\:216$ ways.

Two vegetables and one Other
. . $(_6C_2)(_9C_1) \:=\:(15)(9) \:=\: 135$ ways.

Three vegetables
. . $_6C_3 \:=\:20$ ways.

Therefore, there are: . $216 + 135 + 20 \:=\:371$ vegetarian pizzas.