1
$\begingroup$

I have a probability/combinatorics question that is giving me trouble specifically I don't understand why there is a need for the denominator $6$ "the different orderings of $3$ people" was the explanation but I'm not sure why that is necessary. I've put the question below and my thought process as well.

Question:

If a committee of $3$ people is to be selected from among $5$ married couples so that the committee does not include two people who are married to each other, how many such committees are possible?

Thought Process:

  1. First person of the $3$ selected can be any one among the $10$ folks $$\frac{10!}{(1!)(9!)}$$

  2. Second Person of the $3$ selected must not be married to the first person so lowers availability to $8$ folks to choose from $$\frac{8!}{((1!)(7!)}$$

  3. Third Person of the $3$ selected must not be married to either the first or second person - lowers availability to $6$ folks to choose from $$\frac{6!}{(1!)(5!)}$$

  4. Number of possible ways a committee can be formed: $10 \times 8 \times 6 = 480$

***So I'm able to get to the $480$ but the answer explanation says that I need to divide this $480$ by $6$ and I don't know why because it sounds illogical to do so. Any help would be amazing.

$\endgroup$
  • $\begingroup$ Please read this tutorial on how to typeset mathematics on this site. $\endgroup$ – N. F. Taussig Apr 24 '18 at 10:26
  • 1
    $\begingroup$ Consider a related but easier problem: how many ways can you choose a committee of $5$ men from the $5$ married couples? Let's do it your way: choose any man, $5$ choices; choose another man, $4$ choices, and so on. So there are $5\times4\times3\times2\times1=520$ ways to choose a committee of $5$ men. Does that seem like the logical answer? Or should we divide by $5!$ and get a final answer of $1$? $\endgroup$ – bof Apr 24 '18 at 10:44
0
$\begingroup$

you can do this way too

trio committee with no couples togeather= (total no. of committees - number of committees with couples togeather)

$$=(^{10} C_{3}-^5 C_{1}.8)=80 ways$$

foot note:

$^5 C_{1}:-$ for choosing $1$ pair or couple(i.e, $2$ peoples) and after we've chosen $2$ peoples we are left with $8$ choices for $3rd$ person to form committee that is why we multiply them to get

$^5 C_{1}.8$

$\endgroup$
  • $\begingroup$ Thank you so much!! This different approach was interesting and let me think of it differently 🙏😃 $\endgroup$ – Omar Apr 24 '18 at 12:06
0
$\begingroup$

Suppose the three people who are selected are Anne, Barbara, and Charles.

There are six orders in which those same three people could be selected:

Anne, Barbara, Charles

Anne, Charles, Barbara

Barbara, Anne, Charles

Barbara, Charles, Anne

Charles, Anne, Barbara

Charles, Barbara, Anne

However, all six choices constitute the same committee. Therefore, you need to divide your answer by the $3! = 6$ orders in which you could obtain the same three people, which yields the answer $$\frac{10 \cdot 8 \cdot 6}{3!} = \frac{10 \cdot 8 \cdot 6}{6} = 10 \cdot 8 = 80$$

Alternate Approach: Choose three of the five couples from which to choose the committee members in $\binom{5}{3}$ ways, then choose one of the two members from each selected couple, which can be done in $2^3$ ways. Thus, the number of possible selections is $$\binom{5}{3} \cdot 2^3 = 10 \cdot 8 = 80$$

$\endgroup$
  • 1
    $\begingroup$ Thank you so much mate this was super super helpful 🙏😃 $\endgroup$ – Omar Apr 24 '18 at 12:05

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.