Question

suppose we want to choose 2 letters, without replacement, from the 4 letters a, b, c, and d. (a) how many ways can this be done, if the order of the choices is relevant? (b) how many ways can this be done, if the order of the choices is not relevant?

Explanation:

Part (a)

Step1: Identify as permutation (order matters)

We use the permutation formula \( P(n, k)=\frac{n!}{(n - k)!} \), where \( n = 4 \) (total letters) and \( k = 2 \) (letters to choose).

Step2: Calculate factorials

\( n!=4!=4\times3\times2\times1 = 24 \), \( (n - k)!=(4 - 2)!=2!=2\times1 = 2 \).

Step3: Apply permutation formula

\( P(4, 2)=\frac{4!}{(4 - 2)!}=\frac{24}{2}=12 \).

Step1: Identify as combination (order doesn't matter)

We use the combination formula \( C(n, k)=\frac{n!}{k!(n - k)!} \), where \( n = 4 \) and \( k = 2 \).

Step2: Calculate factorials

\( n!=4!=24 \), \( k!=2!=2 \), \( (n - k)!=(4 - 2)!=2!=2 \).

Step3: Apply combination formula

\( C(4, 2)=\frac{4!}{2!(4 - 2)!}=\frac{24}{2\times2}=6 \).

Answer:

12

Part (b)