Jayvion Caldwell

2022-07-17

Discrete Math Combination
A committee of 12 is to be selected from 10 men and 10 women. In how many ways can the selection be carried out if:
(a) there are no restrictions?
(b) there must be six men and six women?
(c) there must be an even number of women?
(d) there must be more women than men?
(e) there must be at least eight men?
Logical Question:
For part (b), the textbook solution gives $\left(\genfrac{}{}{0}{}{10}{6}\right)×\left(\genfrac{}{}{0}{}{10}{6}\right)$. Why not add up the two possibilities for men and women arrangements ?

gutsy8v

Expert

Step 1
You have $\left(\genfrac{}{}{0}{}{10}{6}\right)$ ways to select 6 women from a group of 10, regardless of which men you picked.
As a consequence, for each group of 6 men you picked, there are $\left(\genfrac{}{}{0}{}{10}{6}\right)$ ways to select the women. But there are $\left(\genfrac{}{}{0}{}{10}{6}\right)$ ways of choosing men, so for each of $\left(\genfrac{}{}{0}{}{10}{6}\right)$ group of men, you can choose $\left(\genfrac{}{}{0}{}{10}{6}\right)$ different groups of women. Hence the multiplication.
Step 2
Maybe a smaller example will be easier to understand.
You're given a menu which has 3 different entrées, 1 main course and 2 deserts. How many different full meals (entrée - main course - desert) can you order?
Let's label the entrées by a,b,c.
If you pick entrée a, you have the choice for two deserts. Hence, two meals with entrée a.
If you pick entrée b, you have the choice for two deserts. Hence, two meals with entrée b.
If you pick entrée c, you have the choice for two deserts. Hence, two meals with entrée c.
Summing it up, you have $3×2$ different meals. You see here where the multiplication comes from: for each entrée, there are 2 meals, so when counting on all entrées you get to multiply the number of entrées to the number of deserts.

Violet Woodward

Expert

Step 1
There are
$\left(\begin{array}{c}10\\ i\end{array}\right)\left(\begin{array}{c}10\\ 10-i\end{array}\right)$
possible committees consisting of i men and $10-i$ women, where i is among 0,...,10. This is true because a male cohort of size i must first be chosen out the the 10 available men, there being
$\left(\begin{array}{c}10\\ i\end{array}\right)$
many ways to do this; moreover, for each choice of male cohort we must choose a female cohort by choosing $10-i$ out of the available 10 females to fill the remaining $10-i$ seats of the committee, and there are
$\left(\begin{array}{c}10\\ 10-i\end{array}\right)$many ways to do this.
Step 2
You multply the binomial coefficents in the last paragraph because of the multiplication principle: if two tasks are to be completed, there are m ways of completing the first task, and for any particular way of completing the first task there are n ways of completing the second task, then there a total of mn ways of completing both tasks. In the last paragraph, the first task would be choosing a male cohort, and the second task would be choosing a female cohort.
Thus if there are no restrictions as to the number of men, women in the committee, then there are
$\sum _{i=0}^{10}\left(\begin{array}{c}10\\ i\end{array}\right)\left(\begin{array}{c}10\\ 10-i\end{array}\right)$possibilities.
The answers to the other parts of this problem can be found by restricting the i in the sum above to have certain properties - for instance, if there must be an even number of males and an even number of females, then i must be even. If there must be at least 8 men, then i can only range from 8 to 10.

Do you have a similar question?