Question

suppose we want to choose 6 colors, without replacement, from 9 distinct colors.
(a) how many ways can this be done, if the order of the choices does not matter?
(b) how many ways can this be done, if the order of the choices matters?

Explanation:

Part (a)

Step1: Identify the problem type

This is a combination problem since the order of choices does not matter. The formula for combinations is \( C(n, k)=\frac{n!}{k!(n - k)!} \), where \( n = 9 \) (total distinct colors) and \( k = 6 \) (colors to choose).

Step2: Substitute values into the formula

Substitute \( n = 9 \) and \( k = 6 \) into the combination formula:

$$ C(9, 6)=\frac{9!}{6!(9 - 6)!}=\frac{9!}{6!3!} $$

We know that \( n!=n\times(n - 1)\times\cdots\times1 \), so \( 9! = 9\times8\times7\times6! \) and \( 3! = 3\times2\times1 \). Substituting these in:

$$ \frac{9\times8\times7\times6!}{6!\times3\times2\times1}=\frac{9\times8\times7}{3\times2\times1} $$

Calculate the numerator and denominator: \( 9\times8\times7 = 504 \) and \( 3\times2\times1 = 6 \). Then \( \frac{504}{6}=84 \).

Step1: Identify the problem type

This is a permutation problem since the order of choices matters. The formula for permutations is \( P(n, k)=\frac{n!}{(n - k)!} \), where \( n = 9 \) and \( k = 6 \).

Step2: Substitute values into the formula

Substitute \( n = 9 \) and \( k = 6 \) into the permutation formula:

$$ P(9, 6)=\frac{9!}{(9 - 6)!}=\frac{9!}{3!} $$

Since \( 9! = 9\times8\times7\times6\times5\times4\times3! \), we can cancel out \( 3! \) from the numerator and denominator:

$$ 9\times8\times7\times6\times5\times4 $$

Calculate the product: \( 9\times8 = 72 \), \( 72\times7 = 504 \), \( 504\times6 = 3024 \), \( 3024\times5 = 15120 \), \( 15120\times4 = 60480 \).

Answer:

(a):
\( 84 \)

Part (b)