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How many 3 topping pizzas can you make with 5 toppings? - Answers
If you can double-up or triple up, then there are (5 x 5 x 5) = 125 possibilities. But actually that is for Permutations(meaning putting on Mushrooms Pepperoni and then Sausage, is a different set of ingredients than Pepperoni Sausage then Mushroom). So actually we want Combinations [order does not matter] See the related link for further explanation. The answer for Combinations with repetition is : n! / (r!(n - r)!), where n = 5 and r = 3, and:5! = 5 x 4 x 3 x 2 x 1 = 120,3! = 3 x 2 x 1 = 6(5 - 3)! = 2! = 2 x 1 = 2, so we have 120 / (6 x 2) = 10.If repetition is allowed, then the formula: (n + r -1)! / (r!(n - 1)!) is used. So we have: (5 + 3 - 1)! / (3!(5-1)!) = 7!/(3! x 4!) = 35 possible.
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How many 3 topping pizzas can you make with 5 toppings? - Answers
If you can double-up or triple up, then there are (5 x 5 x 5) = 125 possibilities. But actually that is for Permutations(meaning putting on Mushrooms Pepperoni and then Sausage, is a different set of ingredients than Pepperoni Sausage then Mushroom). So actually we want Combinations [order does not matter] See the related link for further explanation. The answer for Combinations with repetition is : n! / (r!(n - r)!), where n = 5 and r = 3, and:5! = 5 x 4 x 3 x 2 x 1 = 120,3! = 3 x 2 x 1 = 6(5 - 3)! = 2! = 2 x 1 = 2, so we have 120 / (6 x 2) = 10.If repetition is allowed, then the formula: (n + r -1)! / (r!(n - 1)!) is used. So we have: (5 + 3 - 1)! / (3!(5-1)!) = 7!/(3! x 4!) = 35 possible.
DuckDuckGo
How many 3 topping pizzas can you make with 5 toppings? - Answers
If you can double-up or triple up, then there are (5 x 5 x 5) = 125 possibilities. But actually that is for Permutations(meaning putting on Mushrooms Pepperoni and then Sausage, is a different set of ingredients than Pepperoni Sausage then Mushroom). So actually we want Combinations [order does not matter] See the related link for further explanation. The answer for Combinations with repetition is : n! / (r!(n - r)!), where n = 5 and r = 3, and:5! = 5 x 4 x 3 x 2 x 1 = 120,3! = 3 x 2 x 1 = 6(5 - 3)! = 2! = 2 x 1 = 2, so we have 120 / (6 x 2) = 10.If repetition is allowed, then the formula: (n + r -1)! / (r!(n - 1)!) is used. So we have: (5 + 3 - 1)! / (3!(5-1)!) = 7!/(3! x 4!) = 35 possible.
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