In how many ways can a committee of 8 members be chosen from 7 males and 9 females if the committee must contain at least 3 males?
Please verify if my solution is correct:
total# of 8 members 16C8
#with NO males 7C8
# with at least 3 males
(7C3 x 4C2)+ (7C4 x 4C1)+(7C5 x9C0)
Thanks in advance
I would like to see, in your interpretation, what each of the expressions mean.
(7C3 x 4C2)
(7C4 x 4C1)
(7C5 x9C0)
(7C3 x4C2) # ways of choosing committee 3 males + 5 females. I made a mistake, I think it should be 9C5 instead of 4C2 => (7C3 x 9C5)
(7C4 x 9C4) choosing 4 males + 4 females (again I made a mistake)
Thanks
ok, that sounds reasonable.
If you like, you can continue with your method, which is
(7C3*9C5)+(7C4*9C4)+(7C5*9C3)+(7C6*9C2)+(7C7*9C1)
or, equivalently,
16C8-(7C0*9C8+7C1*9C7+7C2*9C6)
which gives (or should give) the same result above.
In doing counting problems, it is always wise to try two different ways of counting the same objective to check the calculations.
To solve this problem, you can use the concept of combinations. You are correct that the total number of ways to choose 8 members from a group of 16 people (7 males and 9 females) is given by the expression 16C8.
Now, let's examine the different cases for the committee composition:
1. No males: You correctly calculate this as 7C8. However, there seems to be a small mistake: choosing 8 members from a group of only 7 males is not possible. In this case, the number of ways would be 0.
2. At least 3 males: You have correctly broken down this case into three sub-cases based on the number of males chosen.
- 3 males and 2 females: The number of ways to choose 3 males from 7 is given by 7C3, and the number of ways to choose 2 females from 9 is given by 9C2. Therefore, the number of ways for this sub-case is 7C3 * 9C2.
- 4 males and 1 female: The number of ways to choose 4 males from 7 is given by 7C4, and the number of ways to choose 1 female from 9 is given by 9C1. Therefore, the number of ways for this sub-case is 7C4 * 9C1.
- 5 males and 0 females: The number of ways to choose 5 males from 7 is given by 7C5. Since there are no females left to choose, the number of ways for this sub-case is 7C5.
To obtain the total number of ways to form the committee with at least 3 males, you need to sum up the results from each sub-case. Therefore, the correct expression would be:
(7C3 * 9C2) + (7C4 * 9C1) + 7C5
You can simplify this expression further, if desired.
Overall, your solution is almost correct, with a slight error in the first sub-case.