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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?

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    $\begingroup$ 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. $\endgroup$
    – lulu
    Jun 1, 2017 at 10:09

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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 $(a_1a_1, ..., a_{10}a_{10})$

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Can she choose the same topping twice? If she can, the answer is $$10\cdot 9/2 + 10 = 55$$

We divide by two (if the toppings are different) because order does not matter, and we sum the 10 choices you have when you use the same topping twice

If she can't choose the same topping twice, then the answer is

$$10\cdot 9 / 2 = \binom {10}2 = 45$$

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