Question

suppose we want to choose 6 objects, without replacement, from 13 distinct objects.
(a) if the order of the choices does not matter, how many ways can this be done?

(b) if the order of the choices matters, how many ways can this be done?

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 \( n = 13 \) (total objects) and \( k=6 \) (objects to choose).

Step2: Substitute values into the formula

Substitute \( n = 13 \) and \( k = 6 \) into the combination formula:
\[

$$\begin{align*} C(13,6)&=\frac{13!}{6!(13 - 6)!}\\ &=\frac{13!}{6!7!}\\ &=\frac{13\times12\times11\times10\times9\times8\times7!}{6\times5\times4\times3\times2\times1\times7!}\\ &=\frac{13\times12\times11\times10\times9\times8}{6\times5\times4\times3\times2\times1}\\ &=\frac{1235520}{720}\\ & = 1716 \end{align*}$$

\]

Step1: Identify the formula for permutations

When the order matters, we use the permutation formula \( P(n,k)=\frac{n!}{(n - k)!} \), where \( n = 13 \) and \( k = 6 \).

Step2: Substitute values into the formula

Substitute \( n=13 \) and \( k = 6 \) into the permutation formula:
\[

$$\begin{align*} P(13,6)&=\frac{13!}{(13 - 6)!}\\ &=\frac{13!}{7!}\\ &=\frac{13\times12\times11\times10\times9\times8\times7!}{7!}\\ &=13\times12\times11\times10\times9\times8\\ &=1235520 \end{align*}$$

\]

Answer:

1716

Part (b)