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Suppose we want to choose 7 letters, without replacement, from 11 distinct letters. (a)If the order of the choices matters, how many ways can this be done? (b)If the order of the choices does not matter, how many ways can this be done?

Sagot :

jacob193

Answer:

(a) [tex]1\, 663\, 200[/tex] ways to choose when the order matters.

(b) [tex]330[/tex] ways to choose when the order is not considered.

Step-by-step explanation:

The permutation formula gives the number of ways to choose and order [tex]k[/tex] items from a total of [tex]n[/tex] items ([tex]n \ge k[/tex]) without replacement:

[tex]\begin{aligned} {}_{n} P_{k} &= \frac{n!}{(n - k)!}\end{aligned}[/tex].

(The numerator is the factorial of [tex]n[/tex], while the denominator is the factorial of [tex](n - k)[/tex].)

The combination formula is for the case where the order within the [tex]k[/tex] selected items does not matter. The combination formula [tex]{}_{n} C_{k}[/tex] for choosing [tex]k[/tex] items out of a total of [tex]n[/tex] without replacement can be derived in the following steps:

  • Using the permutation formula, find the number of ways to select these [tex]k[/tex] items with ordering:  [tex]{}_{n} P_{k}[/tex].
  • Divide the number of ways to select these [tex]k[/tex] items with ordering by the number of possible ordering within [tex]k[/tex] distinct items, [tex]k![/tex].

In other words:

[tex]\begin{aligned} {}_{n} C_{k} &= \frac{{}_{n} P_{k}}{k!} && \genfrac{}{}{0em}{}{(\text{number of ordered choices})}{(\text{number of orderings within $k$ distinct items})} \\ &= \frac{n!}{(n - k)!\, (k)!}\end{aligned}[/tex].

In this question, [tex]n = 11[/tex] while [tex]k = 7[/tex].

The number of ways to choose when ordering matters (permutation) would be:

[tex]\begin{aligned} {}_{n} P_{k} &= \frac{n!}{(n - k)!} = \frac{11!}{(11 - 7)!} = 1\, 663\, 200\end{aligned}[/tex].

The number of ways to choose without considering ordering within the [tex]k[/tex] selected items (combination) would be:

[tex]\begin{aligned} {}_{n} C_{k} &= \frac{n!}{(n - k)!\, (k)!} = \frac{11!}{(11 - 7)!\, (7!)} = 330\end{aligned}[/tex].

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