1996 AIME Problems/Problem 12

Problem

For each permutation $a_1,a_2,a_3,\cdots,a_{10}$ of the integers $1,2,3,\cdots,10$, form the sum

\[|a_1-a_2|+|a_3-a_4|+|a_5-a_6|+|a_7-a_8|+|a_9-a_{10}|.\]

The average value of all such sums can be written in the form $\dfrac{p}{q}$, where $p$ and $q$ are relatively prime positive integers. Find $p+q$.

Solution

Solution 1

Because of symmetry, we may find all the possible values for $|a_n - a_{n - 1}|$ and multiply by the number of times this value appears. Each occurs $5 \cdot 8!$, because if you fix $a_n$ and $a_{n + 1}$ there are still $8!$ spots for the others and you can do this $5$ times because there are $5$ places $a_n$ and $a_{n + 1}$ can be.

To find all possible values for $|a_n - a_{n - 1}|$ we have to compute \begin{eqnarray*} |1 - 10| + |1 - 9| + \ldots + |1 - 2|\\ + |2 - 10| + \ldots + |2 - 3| + |2 - 1|\\ + \ldots\\ + |10 - 9| \end{eqnarray*}

This is equivalent to

\[2\sum\limits_{k = 1}^{9}\sum\limits_{j = 1}^{k}j = 330\]

The total number of permutations is $10!$, so the average value is $\frac {330 \cdot  8!  \cdot  5}{10!} = \frac {55}{3}$, and $m+n = \boxed{058}$.

Solution 2

Without loss of generality, let $a_1 > a_2;\, a_3 > a_4;\, \ldots ;\, a_9 > a_{10}$. We may do this because all sums obtained from these paired sequences are also obtained in another $2^5-1$ ways by permuting the adjacent terms $\{a_1,a_2\},\{a_3,a_4\}, \cdots , \{a_9, a_{10}\}$, and thus are canceled when the average is taken.

So now we only have to form the sum $S= (a_1 + a_3 + a_5 + a_7 + a_9) - (a_2 + a_4 + a_6 + a_8 + a_{10})$. Due to the symmetry of this situation, we only need to compute the expected value of the result. $10$ must always be the greatest number in its pair; $9$ will be the greater number in its pair $\frac{8}{9}$ of the time and the lesser number $\frac 19$ of the time; $8$ will be the greater number in its pair $\frac 79$ of the time and the lesser $\frac 29$ of the time; and so forth. Each number either adds or subtracts from the sum depending upon whether it is one of the five greater or five lesser numbers in the pairs, respectively. Thus

\begin{align*} \overline{S} &= 10 + \left(\frac{8}{9} \cdot 9\right) - \left(\frac{1}{9} \cdot 9\right) + \left(\frac{7}{9} \cdot 8\right) - \left(\frac 29 \cdot 8\right) + \cdots + \left(\frac{1}{9} \cdot 2\right) - \left(\frac{8}{9} \cdot 2\right) - 1 \\ &= \frac{9 \cdot 10 + 7 \cdot 9 + 5 \cdot 8 + 3 \cdot 7 + 1 \cdot 6 - 1 \cdot 5 - 3 \cdot 4 - 5 \cdot 3 - 7 \cdot 2 - 9 \cdot 1}{9} \\ &= \frac{55}{3} \end{align*}

And the answer is $m+n = \boxed{058}$.

Solution 3

Similar to Solution 1, we can find the average value of $|a_2 - a_1|$, and multiply this by 5 due to symmetry. And again due to symmetry, we can arbitrarily choose $a_2 > a_1$. Thus there are $\binom{10}{2} = 45$ ways to pick the two values of $a_2$ and $a_1$ from the set $\{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$ such that $a_2 > a_1$. First fix $a_2 = 10$, and vary $a_1$ from $1$ to $9$. Then fix $a_2 = 9$, and vary $a_1$ from $1$ to $8$. Continue, and you find that the sum of these $45$ ways to pick $|a_2 - a_1|$ is:

$\sum\limits_{k = 1}^{9}\sum\limits_{j = 1}^{k}j = 45+36+28+21+15+10+6+3+1 = 165$.

Thus, each term contributes on average $\frac{165}{45}$, and the sum will be five times this, or $\frac{165}{9} = \frac{55}{3}$.

The final answer is $m+n = \boxed{058}$.

Solution 4(Expected Value)

We use expected value of one of the sums, say $|a_2 - a_1|$, since all five are similar, so the average or expected value of the sum is just 5 times the expected value of one of them.

To pick two random expected numbers from 1 to 10, we use a well known expected value trick by tying the two ends 1 and 10 with an extra number 0.* This gives 11 spaces, and we distribute 3 gaps evenly to choose our two numbers. We get $\frac{11}{3}$ and $\frac{22}{3}$ giving an absolute difference of $\frac{11}{3}$. Therefore, the average/expected value of the sum of the five would be $\frac{55}{3}$. This gives $55+3=\boxed{058}$.

  • This is like a bendable stick that has equally spaced marks "end-1-2-3-4-5-6-7-8-9-10-end", and the ends are bent together giving a circle with 11 equally spaced marks(the ends produce 1 mark together).

See also

1996 AIME (ProblemsAnswer KeyResources)
Preceded by
Problem 11
Followed by
Problem 13
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

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