Questions: Suppose we want to choose 7 letters, without replacement, from 9 distinct letters. (a) How many ways can this be done, if the order of the choices matters? (b) How many ways can this be done, if the order of the choices does not matter?

Solution Steps

To solve these problems, we need to consider permutations and combinations:

(a) When the order of the choices matters, we use permutations. The number of permutations of choosing 7 letters from 9 distinct letters is given by the formula for permutations: \( P(n, r) = \frac{n!}{(n-r)!} \).

(b) When the order of the choices does not matter, we use combinations. The number of combinations of choosing 7 letters from 9 distinct letters is given by the formula for combinations: \( C(n, r) = \frac{n!}{r!(n-r)!} \).

To find the number of ways to choose 7 letters from 9 distinct letters when the order of choices matters, we use the permutation formula:

\[ P(n, r) = \frac{n!}{(n-r)!} \]

Substituting \( n = 9 \) and \( r = 7 \):

\[ P(9, 7) = \frac{9!}{(9-7)!} = \frac{9!}{2!} = \frac{9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3}{1} = 181440 \]

To find the number of ways to choose 7 letters from 9 distinct letters when the order of choices does not matter, we use the combination formula:

\[ C(n, r) = \frac{n!}{r!(n-r)!} \]

Substituting \( n = 9 \) and \( r = 7 \):

\[ C(9, 7) = \frac{9!}{7! \cdot 2!} = \frac{9 \times 8}{2 \times 1} = 36 \]

Final Answer