A committee of 6 people is chosen from 7 women and 7 men how many different committees are possible that consist of 3 women and 3 men?



Answer :

Answer:

[tex]1225[/tex].

Step-by-step explanation:

The combination formula gives number of ways to choose [tex]r[/tex] objects out of a collection of [tex]n[/tex] total objects is:

[tex]\displaystyle \genfrac{(}{)}{0}{}{n}{k} = \frac{n!}{k!\, (n - k)!}[/tex].

In this question, note that the choice of the [tex]3[/tex] women is independent from the choice of the [tex]3[/tex] men. Hence, start by finding the number of ways to choose women and men separately using the combination formula.

The number of ways to choose [tex]3[/tex] women ([tex]k = 3[/tex]) from the group of [tex]7[/tex] ([tex]n = 7[/tex]) is:

[tex]\displaystyle \genfrac{(}{)}{0}{}{7}{3} = \frac{7!}{3!\, (7 - 3)!} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35[/tex].

In other words, there are [tex]35[/tex] different ways to select the [tex]3[/tex] women out of the group of [tex]7[/tex] women. Similarly, there would be [tex]35[/tex] different ways to choose [tex]3[/tex] men out of a group of [tex]7[/tex] men.

The choice of the [tex]3[/tex] women does not depend on the choice of the [tex]3[/tex] men. Hence, for each of the [tex]35\![/tex] ways to choose [tex]3[/tex] women from the group of women, there would be [tex]35[/tex] unique ways to choose [tex]3[/tex] men from the group of men.

Overall, the unique number of ways to choose [tex]3[/tex] women and [tex]3[/tex] men would be the product of the two [tex]35 \times 35 = 1225[/tex].

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