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Suppose we want to choose 6 letters, without replacement, from 9 distinct letters.
(a) If the order of the choices is not relevant, how many ways can this be done?

◻
(b) If the order of the choices is relevant, how many ways can this be done?

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Suppose we want to choose $6$ letters, without replacement, from $9$ distinct letters. $\newline$ (a) If the order of the choices is not relevant, how many ways can this be done? $\newline$ $\square$ $\newline$ (b) If the order of the choices is relevant, how many ways can this be done? $\newline$ $\square$

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Experimental probability

Full solution

Q. Suppose we want to choose

6

letters, without replacement, from

9

distinct letters.

\newline

(a) If the order of the choices is not relevant, how many ways can this be done?

\newline

\square

\newline

(b) If the order of the choices is relevant, how many ways can this be done?

\newline

\square

Combination Formula: To find the number of ways to choose $6$ letters from $9$ without considering the order (combinations), we use the combination formula: $\newline$ $\text{Number of combinations} = \binom{n}{r} = \frac{n!}{r!(n-r)!}$ $\newline$ where $n = 9$ and $r = 6$ . $\newline$ $\binom{9}{6} = \frac{9!}{6! \cdot (9-6)!} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84$
Calculate Combinations: For the ordered selections (permutations), we use the permutation formula: $\newline$ $\text{Number of permutations} = \frac{n!}{(n-r)!}$ $\newline$ where $n = 9$ and $r = 6$ . $\newline$ $\frac{9!}{(9-6)!} = \frac{9!}{3!} = 9 \times 8 \times 7 \times 6 \times 5 \times 4 = 60480$

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