# From a class of 20 females and 10 males, four students are to be chosen for a committee. How many different ways can the students be chosen such that there would be at least one female?

There are 27,195 ways to choose a committee of four people with at least one female from a class of 20 females and 10 males.

(1) "At least one" female means that the committee can have one, two, three, or four females. Thus we can add the number of each...

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There are 27,195 ways to choose a committee of four people with at least one female from a class of 20 females and 10 males.

(1) "At least one" female means that the committee can have one, two, three, or four females. Thus we can add the number of each of these types of committee together.

1 female: We will select 1 female from 10 possible and 3 males from 10 possible. Order does not matter, so we use combinations:

`_(10)C_3 * _(20)C_1=2400`

**Note that we multiply here since we have two events—this is the multiplication principle.

2 females: Select 2 females and 2 males:

`_10C_2 * _20C_2 = 8550`

3 females: Choose 3 females and 1 male:

`_10C_1 * _20C_3 = 11400`

4 females: `_20C_4=4845`

Then 2400+8550+11400+4845=27195

**Note that we added the four possible different outcomes. One way to remember is to multiply if you say "and" (3 females and 1 male) and to add when you say "or" (1 female or 2 females).

(2) Another approach, often easier when dealing with less than, is to use the complement. The complement of "at least one female" is "no females." If we add all the ways event E can occur to all the ways "not E" can occur, we get the total number of possibilities (or the size of the sample space).

** The notation for the complement varies: C(E), E', ~E, and so on.

So the total number of outcomes is `_30C_4=27405` . The size of the complement of at least one female is the number of ways to choose no females (or all males), which is `_10C_4=210`

So E+E'=total ==> total -E'=E or 27405-210=27195, as before.

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