Question

suppose we want to choose 5 colors, without replacement, from 14 distinct colors.
(a) how many ways can this be done, if the order of the choices is not taken into consideration?
(b) how many ways can this be done, if the order of the choices is taken into consideration?

Explanation:

Part (a)

Step1: Identify the formula for combinations

When the order does not matter, we use the combination formula \( C(n, k)=\frac{n!}{k!(n - k)!} \), where \( n = 14 \) (total number of colors) and \( k = 5 \) (number of colors to choose).

Step2: Substitute the values into the formula

First, calculate the factorials. \( n!=14! = 14\times13\times12\times11\times10\times9\times8\times7\times6\times5\times4\times3\times2\times1 \), \( k!=5! = 5\times4\times3\times2\times1 \), and \( (n - k)!=(14 - 5)!=9! = 9\times8\times7\times6\times5\times4\times3\times2\times1 \).
Then, \( C(14, 5)=\frac{14!}{5!(14 - 5)!}=\frac{14\times13\times12\times11\times10\times9!}{5!\times9!} \). The \( 9! \) terms cancel out.
Now, calculate \( 5! = 120 \), and \( 14\times13\times12\times11\times10=240240 \). Then, \( \frac{240240}{120}=2002 \).

Step1: Identify the formula for permutations

When the order matters, we use the permutation formula \( P(n, k)=\frac{n!}{(n - k)!} \), where \( n = 14 \) and \( k = 5 \).

Step2: Substitute the values into the formula

\( P(14, 5)=\frac{14!}{(14 - 5)!}=\frac{14!}{9!} \). Since \( 14!=14\times13\times12\times11\times10\times9! \), the \( 9! \) terms cancel out.
So, \( 14\times13\times12\times11\times10 = 240240 \).

Answer:

2002

Part (b)