$10 Large Pizza

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.

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!).

1. A pizza with all 3 different toppings

2. A pizza with 2 different toppings

3. A pizza with only 1 topping

4. A pizza with no toppings (Horrified Gasp)

So, the total number of combinations ends up being: 11C0 + 11C1 + 11C2 + 11C3 = 232

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

• Three toppings: From $11$ toppings, there are ${11 \choose 3}$ ways.
• Two toppings: From $11$ toppings, there are ${11 \choose 2}$ ways.
• One topping: From $11$ toppings, there are ${11 \choose 1}$ ways.
• Zero toppings: From $11$ toppings, there are ${11 \choose 0}$ ways.

Thus, our final answer is

${11 \choose 3} + {11 \choose 2} + {11 \choose 1} + {11 \choose 0} = \boxed{232}$

3c0+3c1+3c2+3c3=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: $C(11,0)+ C(11,1)+C(11,2)+C(11,3)=1+11+66+165=232$