# Difference between revisions of "2016 AIME II Problems/Problem 13"

Beatrix is going to place six rooks on a $6 \times 6$ chessboard where both the rows and columns are labeled $1$ to $6$; the rooks are placed so that no two rooks are in the same row or the same column. The $value$ of a square is the sum of its row number and column number. The $score$ of an arrangement of rooks is the least value of any occupied square.The average score over all valid configurations is $\frac{p}{q}$, where $p$ and $q$ are relatively prime positive integers. Find $p+q$.

## Solution 1

We casework to find the number of ways to get each possible score. Note that the lowest possible score is $2$ and the highest possible score is $7$. Let the bijective function $f(x)=\{1,2,3,4,5,6\} \to \{1,2,3,4,5,6\}$ denote the row number of the rook for the corresponding column number.

• For a score of $2$, we must have $f(1)=1$, and we can arrange the rest of the function however we want, so there are $5!=120$ ways.
• For a score of $3$, we must have either $f(1)=2$ or $f(2)=1$, and we can arrange the rest of the rooks however we want, so by PIE the number of ways is $5!+5!-4!=216$.
• For a score of $4$, we must have $f(1)=3$, $f(2)=2$, or $f(3)=1$. If $f(1)=3$, we just don't want $f(2)=1$, if $f(2)=2$, we don't want $f(1)=1$, or if $f(3)=1$, we don't want $f(1)=2$, otherwise we can arrange the function however we like. If at least $2$ of the values rooks have a value of $4$, we can arange the rest of the rooks however we like, so there are $3(5!-4!)-3(4!)+3!=222$ by PIE.
• If the score is $5$, then we have either $f(4)=1$, $f(3)=2$, $f(2)=3$, or $f(1)=4$. If we have the first case, we don't want $f(2)=2$, $f(1)=2$, or $f(1)=3$, so by PIE the number of bad cases is $4!+4!-3!=42$. If we have the second case, then we don't want $f(2)=1$, $f(1)=1$, or $f(1)=3$, so similarly there are $42$ bad cases. Therefore, there are a total of $54$ good cases for each one. The number of ways to get $f(1)=1, f(1)=4$ is $4!-3!$ because we don't want $f(2)=2$, the number of ways to get $f(4)=1, f(2)=3$ is $4!-3!$ ways because we don't want $f(1)=2$, the number of ways to get $f(2)=3, f(3)=2$ is $4!-3!$ ways because we don't want $f(1)=1$, and the number of ways to get $f(1)=4, f(2)=3$ is $4!-3!$ ways because we don't want $f(3)=1$. The number of ways to get at least $3$ cases satisfied is $6! \cdot 4$ because we can arrange the remaining rooks however we like, and the number of ways to get all $4$ cases satisfied is $2!=2$ ways because we can arrange the remaining rooks however we like, so by PIE we have $54 \cdot 4-18 \cdot 6 + 6! \cdot 4-2!=130$ ways to get a score of $5$.
• The only way to get a score of $7$ is to have all the rooks run on the antidiagonal. Therefore, the number of ways to get a sum of $6$ is $6!-120-216-222-130-1=31$.

Thus, the expected sum is $\dfrac{120 \cdot 2 + 216 \cdot 3 + 222 \cdot 4 + 130 \cdot 5 + 31 \cdot 6 + 1 \cdot 7}{720}= \dfrac{2619}{720}=\dfrac{291}{80}$, so the desired answer is $291+80=\boxed{371}$.

## Solution 2

If the score is $n+1$, then one of the rooks must appear in the $n$th antidiagonal, and this is the first antidiagonal in which a rook can appear. To demonstrate this, we draw the following diagram when $n=4$.

$[asy] for (int i=0;i<7;++i) {draw((0,10*i)--(60,10*i));draw((10*i,0)--(10*i,60));} path x=(1,1)--(9,9),y=(1,9)--(9,1); for (int i=0;i<3;++i) {for (int j=3+i;j<6;++j) {draw(shift(10*i,10*j)*x,linewidth(1));draw(shift(10*i,10*j)*y,linewidth(1));}} for (int i=0;i<4;++i) {filldraw((10*i,20+10*i)--(10*i+10,20+10*i)--(10*i+10,20+10*i+10)--(10*i,20+10*i+10)--cycle,lightgray);} [/asy]$ We first count the number of arrangements that avoid the squares above the $n$th diagonal, and then we subtract from these the number of arrangements that avoid all squares above the $(n+1)$th diagonal. In the first column, there are $7-n$ rows in which to place the rook. In the second column, there is one more possible row, but one of the rows is used up by the rook in the first column, hence there are still $7-n$ places to place the rook. This pattern continues through the $n$th column, so there are $(7-n)^n$ ways to place the first $n$ rooks while avoiding the crossed out squares. We can similarly compute that there are $(6-n)^n$ ways to place the rooks in the first $n$ columns that avoid both the crossed out and shaded squares. Therefore, there are $(7-n)^n-(6-n)^n$ ways to place the first $n$ rooks such that at least one of them appears in a shaded square.

After this, there are $(6-n)$ rows and $(6-n)$ columns in which to place the remaining rooks, and we can do this in $(6-n)!$ ways. Hence the number of arrangements with a score of $n$ is $((7-n)^n-(6-n)^n)\cdot(6-n)!$. We also know that $n$ can range from from $1$ to $6$, so the average score is given by

$$\frac{2\cdot(6^1-5^1)\cdot5!+3\cdot(5^2-4^2)\cdot4!+4\cdot(4^3-3^3)\cdot3!+5\cdot(3^4-2^4)\cdot 2!+6\cdot(2^5-1^5)\cdot 1!+7\cdot(1^6-0^6)\cdot 0!}{6!}=\frac{291}{80}.$$ Thus the answer is $\boxed{371}$.