Matching problem question Ten cards are numbered $1$ to $10$ are dealt without replacement one at a time. A match in position $j$ is said to occur if the $j$th card is numbered $j$. Find the probability that there are matches in exactly four of the even numbered positions and in exactly three of the odd numbered positions. 
If $E_1 =$ (matches in $4$ of even # positions), $E_2 =$ (matches in $3$ odd # positions), we know that the $P$(match) $= 1/10$. So $$P(E_1) = {5 \choose 4}\left(\frac{1}{10}\right)^4\left(\frac{9}{10}\right)$$ and $$P(E_2) = {4 \choose 3}\left(\frac{1}{10}\right)^3\left(\frac{9}{10}\right),$$ and the probability of both is product of the probability of the two events. Is this correct?
 A: There are $10!$ permutations of the $10$ cards, with all permutations equally likely.
We ask how many permutations there are with $4$ matches in even-numbered positions, and $3$ matches in odd-numbered positions. These positions can be chosen in $$\binom{5}{4}\binom{5}{3}$$
ways.  
For each such choice, there are $3$ cards left, and their corresponding positions are open.  So we ask how many permutations of $\{a,b,c\}$ there are that have no match. We can list them explicitly: $b, c, a$ and $c, a, b$. Thus our probability is
$$\frac{\binom{5}{4}\binom{5}{3}(2)}{10!}.$$
Comment: The following problem is called the problem of Derangements.  We permute the integers from $1$ to $n$.  How many permutations have no matches?  There are many interesting approaches.  In our problem, we bumped into the problem of counting the number of derangements of $3$ objects. We certainly don't need to consult a book on combinatorics to do that!  But the subject is very interesting, and I could not resist the opportunity to mention it.
A: You can’t simply multiply the probabilities of $E_1$ and $E_2$, because what happens in the even-numbered positions isn’t independent of what happens in the odd-numbered positions. For instance, it’s clearly possible to get matches in exactly four of the even-numbered positions, so $P(E_1)\ne 0$, but if you have matches in all five odd-numbered positions, there is absolutely no way for $E_1$ to occur: you can’t have exactly $9$ matches.
Your calculations of $P(E_1)$ and $P(E_2)$ are also incorrect. Let’s look at $E_1$. There are $\binom54=5$ ways to choose which four of the even-numbered positions will be matches. There are $5$ ways to choose which odd-numbered card will fill the other even-numbered position, and there are $5!$ ways to permute the remaining $5$ cards among the $5$ odd-numbered positions. Thus, there are $5\cdot 5\ \cdot 5! = 25\cdot 120 = 3000$ permutations of the deck that result in $E_1$, and therefore $$P(E_1) = \frac{3000}{10!} = \frac{5}{6048},$$ since there are $10!$ different ways to deal out the cards, all of them equally likely. 
If $n$ is the number of ways to get matches in exactly four of the even-numbered positions and three of the odd-numbered positions, the desired probability is $\dfrac{n}{10!}$, so let’s calculate $n$.
There are $\binom54 = 5$ ways to choose which four of the five even-numbered positions will be matches, and there are $\binom53 = 10$ ways to choose which three of the five odd-numbered positions will be matches, so there are $5\cdot 10 = 50$ ways to specify the seven matching positions.
This leaves three positions, say $p_1,p_2$, and $p_3$, and the three cards numbered $p_1,p_2$, and $p_3$. These three cards will end up in these three positions, so we have to make sure that none of them ends up in the ‘right’ position. There are $3! = 6$ possible permutations of cards $p_1,p_2$, and $p_3$ in positions $p_1,p_2$, and $p_3$, and it’s easy to see that only the permutations $p_3,p_1,p_2$ and $p_2,p_3,p_1$ avoid a match. Thus, for each of the $50$ ways to specify the seven matching positions, there are $2$ ways to fill in the other three positions without getting another match. This means that there are just $50\cdot 2 = 100$ permutations of the deck that produce the desired set of matches: $n = 100$, and the desired probability is $$\frac{100}{10!} = \frac{1}{36288}.$$
