SOLUTION: A pizza shop runs a special where you can buy a large pizza with one cheese, one vegetable and one meat for $9.00. You have a choice of 7 cheeses, 11 vegetables and 6 meats. Additi

Question 450504: A pizza shop runs a special where you can buy a large pizza with one cheese, one vegetable and one meat for $9.00. You have a choice of 7 cheeses, 11 vegetables and 6 meats. Additionally you have a choice of 3 crusts and 2 sauces. How many different variations of the pizza special are possible?
Please,I need help. would appreciate it.
Found 2 solutions by sudhanshu_kmr, solver15:
Answer by sudhanshu_kmr(1152) (Show Source): You can put this solution on YOUR website!

number of possible variations of the pizza = 7*11*6*3*2
= 2772

if any doubt, feel free to contact..

Answer by solver15(27) (Show Source): You can put this solution on YOUR website!
so they gave the choice of one cheese, one vegetable and one meat for $9.00. all oyu have to do is use the permutation formula to find the probabilities.
P(11,1).P(7,1).P(6,1)/P(24,3).You have a choice of 7 cheeses, 11 vegetables and 6 meats. so it will be 11 chose one, 7 chose one, and 6 chose one. and i got P(24,3), 24 by the ones they told you chose from that is 11+7+6=24 and i got 3 was 1+1+1=3.
so P(11,1).P(7,1).P(6,1)/P(24,3) and the formula for permutation is n!/(n-r)! or you can just plug it in your calculator.
=11.7.6/12144
=462/12144
=7/184
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