Question

suppose we want to choose 5 letters, without replacement, from 8 distinct letters.
(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

This is a permutation problem since the order of choices is relevant. The formula for permutations of \( n \) objects taken \( r \) at a time is \( P(n, r)=\frac{n!}{(n - r)!} \), where \( n = 8 \) (total distinct letters) and \( r = 5 \) (letters to choose).

Step2: Substitute values into the formula

Substitute \( n = 8 \) and \( r = 5 \) into \( P(8, 5)=\frac{8!}{(8 - 5)!}=\frac{8!}{3!} \).

Step3: Calculate the factorials

We know that \( n!=n\times(n - 1)\times\cdots\times1 \), so \( 8! = 8\times7\times6\times5\times4\times3! \) and \( 3! = 3\times2\times1 \). Then \( \frac{8!}{3!}=\frac{8\times7\times6\times5\times4\times3!}{3!}=8\times7\times6\times5\times4 \).

Step4: Compute the result

Calculate \( 8\times7 = 56 \), \( 56\times6 = 336 \), \( 336\times5 = 1680 \), \( 1680\times4 = 6720 \).

Step1: Identify the problem type

This is a combination problem since the order of choices is not relevant. The formula for combinations of \( n \) objects taken \( r \) at a time is \( C(n, r)=\frac{n!}{r!(n - r)!} \), where \( n = 8 \) and \( r = 5 \).

Step2: Substitute values into the formula

Substitute \( n = 8 \) and \( r = 5 \) into \( C(8, 5)=\frac{8!}{5!(8 - 5)!}=\frac{8!}{5!3!} \).

Step3: Calculate the factorials

We know that \( 8! = 8\times7\times6\times5! \) and \( 3! = 3\times2\times1 \). Then \( \frac{8!}{5!3!}=\frac{8\times7\times6\times5!}{5!\times3\times2\times1} \).

Step4: Simplify the expression

Cancel out \( 5! \) from the numerator and denominator, then calculate \( \frac{8\times7\times6}{3\times2\times1} \). First, \( 8\times7\times6 = 336 \) and \( 3\times2\times1 = 6 \), so \( \frac{336}{6}=56 \).

Answer:

\( 6720 \)

Part (b)