$4$ people are selected randomly out of six married couples. Then the probability that no husband and wife are selected together.

I wrote answer as $\frac{C(12,1) \cdot C(10,1) \cdot C(8,1) \cdot C(6,1)}{C(12,4)}$

Because first we select any one person out of $12$, now the better half of selected person cannot be selected. Hence, we are left with $10$ options and so on.

What is wrong in my approach? Answer provided in book is $\frac{C(12,1) \cdot C(10,1) \cdot C(8,1) }{3! \cdot C(12,4)}$

When I solved the question alternate approach i.e. Total cases-(Cases when exactly one couple is selected+Cases when exactly two couple is selected) then I got the answer but could someone please tell me what is wrong in first approach?

  • 2
    $\begingroup$ Your mistake in the first attempt was that the way you counted it, you ordered the four people you chose, but the bottom is unordered, causing an overcounting by a factor of $4!$. That said, that doesn't explain the textbook's answer. Could you explain how you counted it the other way? When I tried it that way, as well as by dividing your answer by $4!$, I got $240/{12\choose 4}$. $\endgroup$ – Kevin Long Oct 25 '17 at 19:55

As Kevin Long stated in the comments, the order in which the people are selected does not matter. Therefore, you must divide your answer by the $4!$ orders in which you could select the same four people.

The answer in the book is wrong. Here are two approaches to the problem.

Method 1: We can choose four of the six couples and one of the two people from each selected couple in $$\binom{6}{4}\binom{2}{1}^4$$ ways. Since there are $\binom{12}{4}$ ways to select four of the twelve people, the probability that no husband and wife are selected together is $$\frac{\dbinom{6}{4}\dbinom{2}{1}}{\dbinom{12}{4}}$$

Method 2: We subtract the probability that at least one couple is selected from $1$.

One couple is selected: There are six ways to select a couple and $\binom{10}{2}$ ways to select two of the remaining ten people. Hence, there are $$\binom{6}{1}\binom{10}{2}$$ selections that include a couple.

However, we have counted selections that include two couples twice, once for each way we could designate one of the couples as the selected couple. Since we only want to count them once, we must subtract them from the total.

Two couples are selected: There are $$\binom{6}{2}$$ ways to select two of the six couples.

Hence, the number of ways to select at least one couple is $$\binom{6}{1}\binom{10}{2} - \binom{6}{2}$$ Therefore, the probability of choosing at least one couple is $$\frac{\dbinom{6}{1}\dbinom{10}{2} - \dbinom{6}{2}}{\dbinom{12}{4}}$$ so the probability of not choosing at least one couple is $$1 - \frac{\dbinom{6}{1}\dbinom{10}{2} - \dbinom{6}{2}}{\dbinom{12}{4}}$$


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