Question

the special at a local pizza place says you can choose any 3 toppings from a list of 5. how many possible combinations are there?
a 8
b 10
c 12
d 15

Explanation:

Step1: Identify the problem type

This is a combination problem where we need to find the number of ways to choose 3 toppings from 5, which is a combination of 5 items taken 3 at a time. The formula for combinations is \( C(n, k)=\frac{n!}{k!(n - k)!} \), where \( n = 5 \) and \( k=3 \).

Step2: Calculate factorials

First, calculate the factorials: \( n!=5!=5\times4\times3\times2\times1 = 120 \), \( k!=3!=3\times2\times1=6 \), and \( (n - k)!=(5 - 3)!=2!=2\times1 = 2 \).

Step3: Substitute into the formula

Substitute these values into the combination formula: \( C(5, 3)=\frac{5!}{3!(5 - 3)!}=\frac{120}{6\times2}=\frac{120}{12} = 10 \). Wait, no, wait, let's recalculate. Wait, \( 5! = 120 \), \( 3! = 6 \), \( 2! = 2 \), so \( \frac{120}{6\times2}=\frac{120}{12}=10 \)? Wait, but wait, actually, the correct calculation: \( C(5,3)=\frac{5\times4\times3!}{3!\times2\times1}=\frac{5\times4}{2\times1}=\frac{20}{2} = 10 \)? Wait, no, that's wrong. Wait, no, \( C(5,3)=\frac{5!}{3!2!}=\frac{5\times4\times3!}{3!\times2\times1}=\frac{5\times4}{2\times1}=10 \)? But wait, the options have 10 as option B, but wait, let's check again. Wait, no, actually, \( C(5,3)=\frac{5!}{3!2!}=\frac{120}{6\times2}=10 \). But wait, the correct answer is 10? Wait, no, wait, maybe I made a mistake. Wait, no, the formula for combinations: \( C(n,k)=\frac{n!}{k!(n - k)!} \). So for \( n = 5 \), \( k = 3 \), \( C(5,3)=\frac{5!}{3!2!}=\frac{120}{6\times2}=10 \). So the number of combinations is 10.

Wait, but wait, the options: A is 8, B is 10, C is 12, D is 15. So the correct answer is B. 10.

Answer:

B. 10