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: ___ ___ ___ ___

**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___ ___14___ ___13___ ___12___

**c. Multiply.**

___15___ × ___14___ × ___13___ × ___12___ 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 × 14 × 13 × 12}{4!} = \dfrac{15 × 14 × 13 × 12}{4 × 3 × 2 × 1} = \dfrac{15 × 2 × 7 × 13 × 12}{12 × 2} = 15 × 7 × 13

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

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

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

The correct answer is choice (B).

*Method 2*

Use the formula.

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

Notice that 15! has 11! as a factor.

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

The correct answer is choice (B).