5
$\begingroup$

The United States Supreme Court consists of 3 women and 5 men. In how many ways can a 4-member committee be formed if each committee must have at least two women?.

I know that we have $^8C_4=70$ combinations.

I'm stuck on how many committees can be formed with at least two women.

Do I get the combination of committees that include all men and subtract with the $70$?

$\endgroup$
  • 4
    $\begingroup$ Why is this top network question? $\endgroup$ – theonlygusti Feb 9 '17 at 8:17
  • $\begingroup$ Simply subtracting the number of committees with only men does not eliminate the possibility that a committee has exactly one woman. You would also have to subtract the number of committees that contain exactly one woman from the total. The answers below should give you some idea of how to do that. $\endgroup$ – N. F. Taussig Feb 9 '17 at 9:51
11
$\begingroup$

Check the following combinations:

$1.$ We choose $2$ women and $2$ men. The number of ways for doing so is: $$\binom {3}{2}\times \binom {5}{2} = 3\times 10 =30$$

$2.$ We choose $3$ women and $1$ man. The number of ways for doing so is: $$\binom {3}{3}\times \binom {5}{1} = 1\times 5 =5$$

$3.$ We can choose all $4$ as women. But there are only $3$ women available, so this is not possible.

Thus, the total number of ways to select equals: $30+5=35$ ways. Hope it helps.

$\endgroup$
  • $\begingroup$ Thank you for the detailed answer, now I get it! $\endgroup$ – Killercamin Feb 9 '17 at 4:29
  • $\begingroup$ How might one generalize this method? For example, to choose a 40-member committee with at least 20 women choosing from 30 women and 50 men, the options would be (20 women, 20 men) + (21 women, 19 men) + (22 women, 18 women)+...+(30 women, 10 men)...is there a more general way to calculate this sum? $\endgroup$ – Agargara Feb 9 '17 at 7:57
3
$\begingroup$

You can select 2 or 3 from the 3 available women and select however many men from the 5 available to complete a committee of 4.

$${^3\mathrm C_2}~{^5\mathrm C_2}+{^3\mathrm C_3}~{^5\mathrm C_1} = 35$$

That is all.

$\endgroup$
0
$\begingroup$

If you have 3 women and must choose atleast two then there are two possibilities 1) either you can choose 2women and 2 men 2) or you can choose 3 women and 1 man No of ways to choose 2 women = 3C2 =3 No, of ways to choose 2 men = 5C2 = 10 No of ways to choose 3 women =1 No.of ways to choose 1 man = 5 Therefore the answer will be (3*10)+(1*5)=35

$\endgroup$
  • 2
    $\begingroup$ Please use LaTeX on Math SE. Read the guide to find out how. $\endgroup$ – user21820 Feb 9 '17 at 9:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.