Question

suppose we want to choose 6 colors, without replacement, from 14 distinct colors.
(a) if the order of the choices is relevant, how many ways can this be done?
(b) if the order of the choices is not relevant, how many ways can this be done?

Explanation:

Part (a)

Step1: Identify the problem type

We need to choose 6 colors from 14 distinct colors without replacement and order is not relevant. This is a combination problem. The formula for combinations is \( C(n, k)=\frac{n!}{k!(n - k)!} \), where \( n = 14 \) (total number of colors) and \( k=6 \) (number of colors to choose).

Step2: Calculate the factorial values

First, calculate \( n!=14! = 14\times13\times12\times11\times10\times9\times8\times7\times6\times5\times4\times3\times2\times1 \), \( k! = 6! = 6\times5\times4\times3\times2\times1 \) and \( (n - k)!=(14 - 6)!=8! = 8\times7\times6\times5\times4\times3\times2\times1 \)

Step3: Substitute into the combination formula

\( C(14,6)=\frac{14!}{6!(14 - 6)!}=\frac{14!}{6!×8!} \)
We can simplify this as \( \frac{14\times13\times12\times11\times10\times9\times8!}{6\times5\times4\times3\times2\times1\times8!} \)
The \( 8! \) terms cancel out. Then \( \frac{14\times13\times12\times11\times10\times9}{6\times5\times4\times3\times2\times1} \)
Calculate the numerator: \( 14\times13 = 182 \), \( 182\times12=2184 \), \( 2184\times11 = 24024 \), \( 24024\times10=240240 \), \( 240240\times9 = 2162160 \)
Calculate the denominator: \( 6\times5 = 30 \), \( 30\times4=120 \), \( 120\times3 = 360 \), \( 360\times2=720 \), \( 720\times1=720 \)
Now divide the numerator by the denominator: \( \frac{2162160}{720}=3003 \)

Part (b)

Step1: Identify the problem type

We need to choose 6 colors from 14 distinct colors without replacement and order is relevant. This is a permutation problem. The formula for permutations is \( P(n,k)=\frac{n!}{(n - k)!} \), where \( n = 14 \) and \( k = 6 \)

Step2: Substitute into the permutation formula

\( P(14,6)=\frac{14!}{(14 - 6)!}=\frac{14!}{8!} \)
Since \( 14! = 14\times13\times12\times11\times10\times9\times8! \), we can rewrite the formula as \( 14\times13\times12\times11\times10\times9 \) (because \( 8! \) in the numerator and denominator cancels out)

Step3: Calculate the result

\( 14\times13 = 182 \), \( 182\times12 = 2184 \), \( 2184\times11=24024 \), \( 24024\times10 = 240240 \), \( 240240\times9=2162160 \)

Answer:

3003