Question

answer the questions below.

(a) from a collection of 50 store customers, 2 are to be chosen to receive a special gift. how many groups of 2 customers are possible?

(b) there are 12 european cities that frank would eventually like to visit. on his next vacation, though, he only has time to visit 4 of the cities: one on monday, one on tuesday, one on wednesday, and one on thursday. he is now trying to make a schedule of which city hell visit on which day. how many different schedules are possible? (assume that he will not visit a city more than once.)

Explanation:

Part (a)

Step1: Identify the problem type

This is a combination problem since the order of choosing the 2 customers does not matter. The formula for combinations is \( C(n, k)=\frac{n!}{k!(n - k)!} \), where \( n = 50 \) (total number of customers) and \( k=2 \) (number of customers to be chosen).

Step2: Calculate the combination

First, calculate the factorials:

• \( n!=50! = 50\times49\times48! \)
• \( k!=2! = 2\times1=2 \)
• \( (n - k)!=(50 - 2)!=48! \)

Substitute into the formula:
\( C(50, 2)=\frac{50!}{2!(50 - 2)!}=\frac{50\times49\times48!}{2\times1\times48!} \)

The \( 48! \) terms cancel out:
\( C(50, 2)=\frac{50\times49}{2\times1}=\frac{2450}{2} = 1225 \)

Step1: Identify the problem type

This is a permutation problem since the order of visiting the cities (which city on which day) matters. The formula for permutations is \( P(n, k)=\frac{n!}{(n - k)!} \), where \( n = 12 \) (total number of cities) and \( k = 4 \) (number of cities to be visited).

Step2: Calculate the permutation

First, calculate the factorials:

• \( n!=12! = 12\times11\times10\times9\times8! \)
• \( (n - k)!=(12 - 4)!=8! \)

Substitute into the formula:
\( P(12, 4)=\frac{12!}{(12 - 4)!}=\frac{12\times11\times10\times9\times8!}{8!} \)

The \( 8! \) terms cancel out:
\( P(12, 4)=12\times11\times10\times9=11880 \)

Answer:

1225

Part (b)