The local pizzeria is advertising a special where you can get a large pizza with up to 3 different toppings for only $10. If the pizzeria offers 11 different toppings, how many different types of pizza can you get in this special deal?
Details and assumptions
A double topping of pepperoni does not qualify for the special $10 price. The 3 toppings must be different.
In America, pizzas are purchased according to what toppings are selected on the pizza, as opposed to a type of pizza (say Meat-Lovers Supreme).
The order of toppings on a pizza does not matter.
You can get a pizza with no toppings.
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Right, missed ${11 \choose 0}$ . Question should be amended.
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Agreed. Why would I want to purchase a pizza with no topping?
Is 11C0 possible,i.e. no topping?
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Yes, a pizza with no topping still counts as a pizza, though the question should have probably made this clearer.
I feel the question needs to clarify that the pizza can have no topping. I answered 231 too.
Congratulations to the creator of this sum... he wins a free" topping less" pizza treat from me....!!!!!!!!!!!!!!!!!!!!!!!! I want my points back
I didn't think that no topping was possible.. silly me, this is math and not real life..
I answered 231, pizza with no topping?? i think the guy who made this problem was just trolling haha
If there can be up to $3$ toppings, there can be $0, 1, 2$ , or $3$ toppings.
Ways with $0$ toppings: $1$ .
Ways with $1$ topping: $11$ .
Ways with $2$ toppings: $\binom{11}{2}=55$ .
Ways with $3$ toppings: $\binom{11}{3}=165$ .
$1+11+55+165=232$ .
Well, given the restrictions (11 types of toppings; up to 3 different toppings), you can have four different types of pizzas (pizzas, not toppings!).
A pizza with all 3 different toppings
A pizza with 2 different toppings
A pizza with only 1 topping
A pizza with no toppings (Horrified Gasp)
So, the total number of combinations ends up being: 11C0 + 11C1 + 11C2 + 11C3 = 232
Horrified Gasp indeed!
We note that a pizza can have either 0, 1, 2, or 3 toppings, chosen from 11, so we merely have to sum the appropriate combinations, such that we obtain:
${11 \choose 0}+{11 \choose 1}+{11 \choose 2}+{11 \choose 3}=\boxed{232}$
The pizza can get from $0$ to $3$ different toppings, of $11$ toppings in total. Obviously, there's only $1$ pizza without toppings, $11$ pizzas of only $1$ topping, $\frac {11!}{2!\cdot9!} = 55$ with $2$ different toppings, and $\frac {11!}{3!\cdot8!} = 165$ of $3$ different toppings. In conclusion, we have $1 + 11 + 55 + 165 = \boxed {232}$ types of pizza.
Since the order of toppings dont matter, so we take here combination to find out the number of ways (not permutation). The pizza ordered can have no topping or $1$ topping or $2$ toppings or $3$ toppings at most so as to avail the special deal. Now, there are $11$ different possible toppings that can be added to the pizza.
Number of different pizzas that can have no topping $=11C0=1$
Number of different pizzas that can have $1$ topping $=11C1=11$
Number of different pizzas that can have $2$ toppings $=11C2=\frac{11\times 10}{2}=11\times 5 = 55$
Number of different pizzas that can have $3$ toppings $=11C3=\frac{11\times 10\times 9}{3\times 2}=11\times 15 = 165$
So, total number of different pizzas that can be bought $=1+11+55+165=\boxed{232}$
The pizza can have
Thus, our final answer is
${11 \choose 3} + {11 \choose 2} + {11 \choose 1} + {11 \choose 0} = \boxed{232}$
You can either choose $0, 1, 2$ or $3$ unique toppings to qualify for the offer. Therefore, the total number of pizzas available is, ${11 \choose 0} + {11 \choose 1} + {11 \choose 2} + {11 \choose 3} = \fbox{232}$
Pizza could have 0 toppings, 1 toppings, 2 toppings or 3 toppings among the 11: $C(11,0)+ C(11,1)+C(11,2)+C(11,3)=1+11+66+165=232$
I hope you meant to say $\binom{11}{2} = 55$ and not $66$ ... looks like you punched in the wrong keys in haste
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Note that the number of ways to top the pizza with $n$ toppings is $\displaystyle\binom{11}{n}$ . Since we want to know the total where $n=0,1,2,3$ , we sum together: $\displaystyle\binom{11}{0}+\binom{11}{1}+\binom{11}{2}+\binom{11}{3}=\boxed{232}$ and we are done.