Question

1. johnny is a very picky eater, so he likes to use a lot of condiments. he has ketchup, salt, pepper, and shredded cheese at his disposal. his mother tells him he may only make two additions to his meal (i.e., he can add condiments only twice, regardless of whether or not he already used them). how many different ways can johnny improve his meal?

6

10

12

16

Explanation:

Step1: Identify the problem type

This is a problem of counting the number of ways to choose 2 condiments with repetition allowed (since he can use the same condiment twice). The formula for combinations with repetition is \( C(n + r - 1, r) \), where \( n \) is the number of types of condiments and \( r \) is the number of selections. Here, \( n = 4 \) (ketchup, salt, pepper, shredded cheese) and \( r = 2 \).

Step2: Apply the formula

First, calculate \( n + r - 1 = 4 + 2 - 1 = 5 \). Then, \( C(5, 2)=\frac{5!}{2!(5 - 2)!}=\frac{5\times4}{2\times1}=10 \). Alternatively, we can list the cases:

• Using the same condiment twice: 4 ways (ketchup - ketchup, salt - salt, pepper - pepper, shredded cheese - shredded cheese).
• Using different condiments: \( C(4, 2)=\frac{4!}{2!(4 - 2)!}=\frac{4\times3}{2\times1}=6 \) ways.
• Total ways: \( 4 + 6 = 10 \).

Answer:

10