Answer:
a lot
960
Explanation:
Since there are 4 woman in total, there are 14-4=10 men.
Part 1
They can choose either 2 women, 3 women, or 4 woman.
If they choose 2 out of the 4 woman, they can have 6 different combinations.
4!/(2!(4-2)!)
=24/(2(2!))
=24/4=6
There are 8 remaining men to be chosen on the committee (because 10-2=8).
There are 10 total men and 8 men need to be chosen out of them, so the number of combinations for men is given below.
10!/(8!(10-8)!)
=10!/(8!(2!))
Instead of multiplying out 1x2x3x4x5x6x7x8x9x10 to find 10!, we can divide 8! from both the numerator and denominator to simplify it. This leaves the numerator with 9x10, which is 90, and the denominator with 2!, which is just 2.
90/2=45
Now we need to multiply 45 (number of combinations to choose 8 men from 10 men) and 6 (number of combinations to choose 2 women from 4 woman).
45x6=270 combinations if 2 woman are chosen
Part 2
Next we need to find out how many combinations there are if they choose 3 women for the committee.
If they choose 3 out of the 4 woman, they can have 4 different combinations.
4!/(3!(4-3)!)
=24/(6(1!))
=24/6=4
There are 7 remaining men to be chosen on the committee (because 10-3=7).
There are 10 total men and 7 men need to be chosen out of them, so the number of combinations for men is given below.
10!/(7!(10-7)!)
=10!/(7!(3!))
Instead of multiplying out 1x2x3x4x5x6x7x8x9x10 to find 10!, we can divide 7! from both the numerator and denominator to simplify it. This leaves the numerator with 8x9x10, which is 720, and the denominator with 3!, which is 6.
720/6=120
Now we need to multiply 120 (number of combinations to choose 7 men from 10 men) and 4 (number of combinations to choose 3 women from 4 woman).
120x4=480 total combinations if 3 women are chosen
Part 3
If 4 women out of 4 women are chosen for the committee, then obviously, there will only be 1 combination for that, since all the women will be chosen.
This leaves us with 10-4=6 spots for men in the committee.
There are 10 total men and 6 men need to be chosen out of them, so the number of combinations for men is given below.
10!/(6!(10-6)!)
=10!/(6!(4!))
Instead of multiplying out 1x2x3x4x5x6x7x8x9x10 to find 10!, we can divide 6! from both the numerator and denominator to simplify it. This leaves the numerator with 7x8x9x10, which is 5040, and the denominator with 4!, which is 24.
5040/24=210
Now we need to multiply 210 (number of combinations to choose 6 men from 10 men) and 1 (number of combinations to choose 4 women from 4 woman).
210x1=210 total combinations if 3 women are chosen
Part 4
Finally, we need to add all the final answers from the previous parts.
270+480+210=960 total combinations if at least 2 women were chosen for the committee
Please let me know if you find any errors