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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 ${10 \choose 6} \times {10 \choose 6}$. Why not add up the two possibilities for men and women arrangements ?

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    I think of (10 choose 6) + (10 choose 6), why this is not possible for this case ? – 2017-02-16
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    The correct question is not "Why not do X?" but "Why do X?". Namely, why do you want to add? If you cannot explain why, then there is a high likelihood it is wrong. – 2017-02-16
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    I am starting to learn discrete math, so at this point asking why I need to do this requires me to provide proof which I don't have the knowledge for yet. Mostly, I am thinking about it out of curiosity or trying to see if other possibilities exist. – 2017-02-16
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    Well mathematically yes to answer "why?" would require a proof, but I was actually asking for your "intuitive reason" for doing so. In mathematics, although a proof is the only means of being sure of the correctness of a claim, it should not be the only source of the conviction. Prior to that should be an intuition for the correctness, which one then tries to translate into a proof. In this case, the intuition is more or less what you wrote in a later comment, which I shall now respond to. – 2017-02-17
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    So after you read my response, a good question for you to answer is: What task that sounds similar to (b) actually has ( 10 choose 6 ) + ( 10 choose 6 ) as the answer? – 2017-02-17
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    Ok I see. If you mean to give an example, then we can say like a mother needs to prepare lunch for the kids. There must be six water bottles and six lunch bowls. – 2017-02-17
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    I don't get your example. Your question (b) has answer ( 10 choose 6 ) **times** ( 10 choose 6 ). I'm asking you what question has answer ( 10 choose 6 ) **plus** ( 10 choose 6 ). – 2017-02-17
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    Oh, my bad, I was just confused when to use adding and multiplication. There is no actual question that has ( 10 choose 6 ) plus ( 10 choose 6 ). Sorry for misunderstanding. – 2017-02-17
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    There **is**. I'm asking you to come up with a question that sounds similar to (b) but **does have** the answer ( 10 choose 6 ) + ( 10 choose 6 ). – 2017-02-18

2 Answers 2

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You have $10 \choose 6$ 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 $10 \choose 6$ ways to select the women. But there are $10 \choose 6$ ways of choosing men, so for each of $10 \choose 6$ group of men, you can choose $10 \choose 6$ different groups of women. Hence the multiplication.

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 $3 \times 2$ 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.

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    Very clear approach to this question. So @Mariuslp, we use the product rule if two groups like men and women are related to each other, but also distinct of each other ? In other words, considering the food example you gave, we can multiply them because they are distinct of each other, but they are related in a common thing which is trying to see how many arrangements of foods are possible. Is my logic right ? – 2017-02-16
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    Sorry for the late reply. Yes, this is correct! As the entrées are independent from the deserts, you can multiply. The same is true with groups of men and women. Of course, you can't multiply anymore when there is some dependence between your groups, such as: "Mary and Tom cannot be picked together". Then you need some refinement. – 2017-02-19
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There are

\begin{equation} \left(\begin{matrix} 10 \\ i \end{matrix}\right)\left(\begin{matrix} 10 \\ 10-i \end{matrix}\right)\end{equation}

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 \begin{equation} \left(\begin{matrix} 10 \\ i \end{matrix}\right)\end{equation} 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 \begin{equation} \left(\begin{matrix} 10 \\ 10-i \end{matrix}\right) \end{equation} many ways to do this.

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

\begin{equation} \sum_{i=0}^{10}\left(\begin{matrix} 10 \\ i \end{matrix}\right)\left(\begin{matrix} 10 \\ 10-i \end{matrix}\right) \end{equation}

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 $10$. Etc

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    Hmm, interesting, so you mean we can use the product rule, when two tasks are distinct from each other ? Like task m and n, thus m * n. – 2017-02-16
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    Because we can choose the # of women and men without them being directly dependent on each other. They are two independent process, but they both share a common thing, finding the number of arrangements, so we multiply them. This is what my logic is telling me. – 2017-02-16
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    @Ciruss: Right, but that we should not think of the two tasks as 'sharing' any common thing. Put differently, you have a big task which is to **first** choose 6 men and **then** choose 6 women. To complete the big task you have to complete the first small task (choose 6 men) and then you reach the point where you then have to complete the second small task (choose 6 women) and the number of ways you can do the second small task is **independent** of the way you did the first small task, so the number of ways you can complete the big task is the product. – 2017-02-17
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    @Ciruss: In contrast, if you have a big task where your first step is to **choose** to do exactly one small task (and none of the others), then clearly the number of ways you can do the big task is the sum of the number of ways for each small task. Do tell me if this makes sense. – 2017-02-17
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    @Ciruss: Note that my above two comments are not a proof! But they should capture the intuitive idea sufficiently to be convincing. A formal statement of this discrete mathematics principle would actually be quite cumbersome, so it's not really worth to go into that at this point. – 2017-02-17
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    Yes your approach makes sense, for example, you need to choose 4 science and math books, but you need to choose at least 2 science books. Then we get (4 choose 2) * (2 choose 2) + (4 choose 3) * (1 choose 1). This example shows that the number of math books depend on the number of science books, so this dependency and relation forces us to use 'rule of product'. On the other hand, the word at least is used, so we have two situations that suits this question, and these two situations can be added up using the 'rule of sum'. – 2017-02-17
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    @Ciruss: Wait, "choose 4 science and maths books" from what? – 2017-02-17
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    By my example, I meant you have total of 4 books, and you need to choose at least 2 science books, and the rest math books. I guess that question shows clearly how both rules of sum and product are used within a question. – 2017-02-17
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    So do you confirm my logic in that example ? – 2017-02-17
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    @Kourosh: You **still** didn't answer my question. Choose **from what**? – 2017-02-18