A committee of 6 is to be chosen from the 28 students in a class. If there are 10 males and 18 females in the class, in how many ways can this be done if there must be at least three females on the committee? A: 339864B: 816720C: 3060D: 18564

Respuesta :

Hello! First, let's write some important information contained in the exercise:

committee = 6 students

class: 28 students:

- 10 males

- 18 females

Let's consider the rule: At least three females must be on the committee, so we have some cases, look:

_F_ * _F_ * _F_ * __ * __ * __

1st option:

3 females and 3 males

_F_ * _F_ * _F_ * _M_ * _M_ * _M_

2nd option:

4 females and 2 males

_F_ * _F_ * _F_ * _F_ * _M_ * _M_

3rd option:

5 females and 1 male

_F_ * _F_ * _F_ * _F_ * _F_ * _M_

4th option:

6 females and 0 male

_F_ * _F_ * _F_ * _F_ * _F_ * _F_

Now, we have to use the formula below and find the number of possible combinations:

[tex]C_{n,p}=\frac{n!}{p!\cdot(n-p)!}[/tex]

Let's calculate each option below:

1st:

3 females:

[tex]C_{18,3}=\frac{18!}{3!\cdot(18-3)!}=\frac{18\cdot17\cdot16\cdot15!}{3\cdot2\cdot1\cdot15!}=\frac{4896}{6}=816[/tex]

3 males:

[tex]C_{10,3}=\frac{10!}{3!\cdot(10-3)!}=\frac{10\cdot9\cdot8\cdot7!}{3\cdot2\cdot1\cdot7!}=\frac{720}{6}=120[/tex]

3 females and 3 males: 816 * 120 = 97920

2nd option:

4 females:

[tex]C_{18,4}=\frac{18!}{4!\cdot(18-4)!}=\frac{18\cdot17\cdot16\cdot15\cdot14!}{4\cdot3\cdot2\cdot1\cdot14!}=\frac{73440}{24}=3060[/tex]

2 males:

[tex]C2=\frac{10!}{2!\cdot(10-2)!}=\frac{10\cdot9\cdot8!}{2\cdot1\cdot8!}=\frac{90}{2}=45[/tex]

4 females and 2 males: 3060* 45 = 137700

3rd option:

5 females:

[tex]C_{18,5}=\frac{18!}{5!\cdot(18-5)!}=\frac{18\cdot17\cdot16\cdot15\cdot14\cdot13!}{5\cdot4\cdot3\cdot2\cdot1\cdot13!}=\frac{1028160}{120}=8568[/tex]

1 male:

[tex]C_{10,1}=\frac{10!}{1!\cdot(10-1)!}=\frac{10!}{1\cdot9!}=\frac{3628800}{362880}=10[/tex]

5 females and 1 male = 8568 * 10 = 85680

4th option:

6 females and 0 male:

[tex]C_{18,6}=\frac{18!}{6!\cdot(18-6)!}=\frac{18\cdot17\cdot16\cdot15\cdot14\cdot13\cdot12!}{6\cdot5\cdot4\cdot3\cdot2\cdot1\cdot12!}=\frac{13366080}{720}=18564[/tex][tex]C_{10,0}=\frac{10!}{0!\cdot(10-0)!}=\frac{10!}{10!}=1[/tex]

6 females and 0 male: 18564 * 1 = 18564

To finish the exercise, we have to sum the four options:

97920 + 137700 + 85680 + 18564 = 339864

So, right answer A: 339864.