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n balls that are numbered $1, 2, \ldots, n$ are randomly distributed into $n$ cells that are also numbered $1, 2, \dots, n$. Every cell can place exactly $1$ ball. Let $X$ be the number of balls that were put into cells with identical numbers. Compute $E(X)$.

So from what I understand so far, I would have to find the general formula of $P(X = k)$ for $k = 1,2,\ldots n$

For $P(X = n)$ it is quite simple which is 1/n! For $P(X = k), 1 \le k < n$, I am stuck as I have only figured out a way of choosing the cells which are identical with the numbers on the ball which is just $n$ choose $k$. What about the rest of the cells? For example, we have $5$ cells. $2$ of the cells are going to be identical with the number on the balls. That is just $5$ choose $2$. What about the other $3$ cells? If using brute force, out of the $6$ ways of putting the balls into the cells, only $2$ would suit our condition. How do we find the general formula for it?

Sorry if my notation is not correct but I hope you understand what I am asking.

TShiong
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Firdaus Hafizi
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1 Answers1

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The Simple Solution Using Linearity of Expectation

In each cell, the expected number of balls with the correct number is $\frac1n$. There are $n$ cells, so by Linearity of Expectation, the expected number is $n\cdot\frac1n=1$.

A More Deranged Answer

The number of Derangements of $n$ items is $$ \mathcal{D}(n)=\sum_{j=0}^n(-1)^j\frac{n!}{j!}\tag{1} $$ The number of ways to have $k$ items in the right place is $$ \binom{n}{k}\mathcal{D}(n-k)\tag{2} $$ that is, the number of ways to choose the $k$ items times the number of derangements of the rest.

Because the following sum counts the total number of permutations of $n$ items, we have $$ \sum_{k=0}^n\binom{n}{k}\mathcal{D}(n-k)=n!\tag{3} $$ Thus, the expected number of balls in the correct cells would be $$ \begin{align} \frac1{n!}\sum_{k=0}^nk\binom{n}{k}\mathcal{D}(n-k) &=\frac1{n!}\sum_{k=1}^nn\binom{n-1}{k-1}\mathcal{D}((n-1)-(k-1))\tag{4}\\ &=\frac1{(n-1)!}(n-1)!\tag{5}\\[9pt] &=1\tag{6} \end{align} $$ Explanation:
$(4)$: $k\binom{n}{k}=n\binom{n-1}{k-1}$
$(5)$: apply $(3)$ to $n-1$
$(6)$: divide

The Variance

Using the same, deranged method as above, we get $$ \begin{align} \frac1{n!}\sum_{k=0}^nk^2\binom{n}{k}\mathcal{D}(n-k) &=\frac1{n!}\sum_{k=0}^n(k(k-1)+k)\binom{n}{k}\mathcal{D}(n-k)\\ &=\frac1{n!}\sum_{k=2}^nn(n-1)\binom{n-2}{k-2}\mathcal{D}((n-2)-(k-2))\\ &+\frac1{n!}\sum_{k=1}^nn\binom{n-1}{k-1}\mathcal{D}((n-1)-(k-1))\\ &=\frac1{(n-2)!}(n-2)!+\frac1{(n-1)!}(n-1)!\\[9pt] &=2\tag{7} \end{align} $$ The Variance is the mean of the square minus the square of the mean. Thus, the variance is $2-1=1$.

robjohn
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