2016 AIME II Problems/Problem 13
Problem
Beatrix is going to place six rooks on a chessboard where both the rows and columns are labeled
to
; the rooks are placed so that no two rooks are in the same row or the same column. The
of a square is the sum of its row number and column number. The
of an arrangement of rooks is the least value of any occupied square.The average score over all valid configurations is
, where
and
are relatively prime positive integers. Find
.
Solution 1
We casework to find the number of ways to get each possible score. Note that the lowest possible score is and the highest possible score is
. Let the bijective function
denote the row number of the rook for the corresponding column number.
- For a score of
, we must have
, and we can arrange the rest of the function however we want, so there are
ways.
- For a score of
, we must have either
or
, and we can arrange the rest of the rooks however we want, so by PIE the number of ways is
.
- For a score of
, we must have
,
, or
. If
, we just don't want
, if
, we don't want
, or if
, we don't want
, otherwise we can arrange the function however we like. If at least
of the values rooks have a value of
, we can arange the rest of the rooks however we like, so there are
by PIE.
- If the score is
, then we have either
,
,
, or
. If we have the first case, we don't want
,
, or
, so by PIE the number of bad cases is
. If we have the second case, then we don't want
,
, or
, so similarly there are
bad cases. Therefore, there are a total of
good cases for each one. The number of ways to get
is
because we don't want
, the number of ways to get
is
ways because we don't want
, the number of ways to get
is
ways because we don't want
, and the number of ways to get
is
ways because we don't want
. The number of ways to get at least
cases satisfied is
because we can arrange the remaining rooks however we like, and the number of ways to get all
cases satisfied is
ways because we can arrange the remaining rooks however we like, so by PIE we have
ways to get a score of
.
- The only way to get a score of
is to have all the rooks run on the antidiagonal. Therefore, the number of ways to get a sum of
is
.
Thus, the expected sum is , so the desired answer is
.
Solution 2
If the score is , then one of the rooks must appear in the
th antidiagonal, and this is the first antidiagonal in which a rook can appear. To demonstrate this, we draw the following diagram when
.
We first count the number of arrangements that avoid the squares above the
th diagonal, and then we subtract from these the number of arrangements that avoid all squares above the
th diagonal. In the first column, there are
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
places to place the rook. This pattern continues through the
th column, so there are
ways to place the first
rooks while avoiding the crossed out squares. We can similarly compute that there are
ways to place the rooks in the first
columns that avoid both the crossed out and shaded squares. Therefore, there are
ways to place the first
rooks such that at least one of them appears in a shaded square.
After this, there are rows and
columns in which to place the remaining rooks, and we can do this in
ways. Hence the number of arrangements with a score of
is
. We also know that
can range from from
to
, so the average score is given by
Thus the answer is
.
Solution 3
So we first count the number of permutations with score . This is obviously
. Then, the number of permutations with score
can also be computed: in the first column, there are five ways to place a rook- anywhere but the place with score
. In the next column, there are
ways to place a rook- anywhere but the one in the same row as the previous row. We can continue this to obtain that the number of permutations with score
is
. Doing the same for scores
,
,
, and
we obtain that these respective numbers are
,
,
,
.
Now, note that if is the number of permutations with score
, then
, where
is the number of permutations with score exactly
. Thus, we can compute the number of permutations with scores
,
, etc as
. We then compute
leading us to the answer of
.
See also
2016 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 12 |
Followed by Problem 14 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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