# probability. how many different choices could she make? is it correct?

A pizza shop offers a selection of $10$ different pizza toppings on its pizzas. Erika orders a pizza with $2$ toppings. How many different choices could she make?

I think that the right answer is $10^2$, but I want to be sure that is correct. Could someone help me?

• No...presumably choosing $A$ and then $B$ is the same as choosing $B$ then $A$. Also, you have to specify whether or not "no topping" or "just $A$" or "double $A$" are options.
– lulu
Jun 1, 2017 at 10:09

The number of ways of choosing two different toppings will be $$9+8+...+1 = 45$$ Since, if our toppings are $a_1, a_2, ...,a_{10}$, then our toppings choices are limited to the following combinations if there is no chance of repeating a topping: $$a_1a_2, a_1a_3, ..., a_1a_{10} \qquad \qquad ...9\ choices \\\qquad a_2a_3, ..., a_2a_{10} \qquad \qquad ...8\ choices \\ . \\. \\. \\\qquad \qquad \quad a_9a_{10}\qquad \qquad ...1\ choice$$ If repeated toppings are allowed, we have an additional 10 choices $(a_1a_1, ..., a_{10}a_{10})$
Can she choose the same topping twice? If she can, the answer is $$10\cdot 9/2 + 10 = 55$$
$$10\cdot 9 / 2 = \binom {10}2 = 45$$