The order of toppings on a pizza does not matter. (Does it matter if pepperoni goes on before mushrooms?) This is a combinations question.

*Method 1*

**1. Start the problem as if it was a permutations problem.**

**a. Figure out how many places there are to fill.**

You are choosing 4 toppings, so there are 4 outcomes:Â Â ___Â Â ___Â Â ___Â Â ___

**b. Figure out how many objects can potentially go into each place.**

There are 15 toppings to choose from. The number of different toppings to choose from decreases after each choice: ___15___Â Â ___14___Â Â ___13___Â Â ___12___

**c. Multiply.**

___15___ Ã— ___14___ Ã— ___13___ Ã— ___12___ Â Â Â Â Â Â Don’t calculate yet. Wait and cancel first.

##### 2. Divide the answer by the factorial of the number of places.

You are eliminating all the repeat pizzas and so you only count the distinct ones.

There are 4 outcomes, so divide by 4!

\dfrac{15 Ã— 14 Ã— 13 Ã— 12}{4!} = \dfrac{15 Ã— 14 Ã— 13 Ã— 12}{4 Ã— 3 Ã— 2 Ã— 1} = \dfrac{15 Ã— 2 Ã— 7 Ã— 13 Ã— 12}{12 Ã— 2} = 15 Ã— 7 Ã— 13

Remember that the GRE is not looking for long calculations. Look at the equation and the answers.

The last digit of 15 Ã— 7 Ã— 13 will be 5, since multiplying the last digits gives you the ones place of the answer. 5 Ã— 7 Ã— 3 = 105, and the only answer choice that ends in 5 is choice (B).

If you do the calculation, it will be 15 Ã— 7 Ã— 13 = 1365.

The correct answer is choice (B).

*Method 2*

Use the formula.

15 *C* 4 = \dfrac{15!}{4! (15 \,-\, 4)!} = \dfrac{15!}{4! Ã— 11!} = \dfrac{15 Ã— 14 Ã— 13 Ã— 12 Ã— 11!}{4 Ã— 3 Ã— 2 Ã— 11!} = \dfrac{15 Ã— 2 Ã— 7 Ã— 13 Ã— 12}{12 Ã— 2}Â = 15 Ã— 7 Ã— 13

Notice that 15! has 11! as a factor.

The last digit of 15 Ã— 7 Ã— 13 will be 5, since multiplying the last digits gives you the ones place of the answer. 5 Ã— 7 Ã— 3 = 105, and the only answer choice that ends in 5 is choice (B).

The correct answer is choice (B).