# University of South Carolina High School Math Contest/1993 Exam/Problem 14

## Problem

How many permutations of 1, 2, 3, 4, 5, 6, 7, 8, 9 have:

• 1 appearing somewhere to the left of 2,
• 3 somewhere to the left of 4, and
• 5 somewhere to the left of 6?

For example, 8 1 5 7 2 3 9 4 6 would be such a permutation. $\mathrm{(A) \ }9\cdot 7! \qquad \mathrm{(B) \ } 8! \qquad \mathrm{(C) \ }5!4! \qquad \mathrm{(D) \ }8!4! \qquad \mathrm{(E) \ }8!+6!+4!$

## Solution

There are $9!$ (the factorial of 9) total permutations of the elements of that set. 1 is to the left of 2 in exactly half of these. 3 is also to the left of 4 in exactly half of the permutations, and 5 is to the left of 6 in exactly half of the permutations. These three events are totally independent of each other, so the number we want to calculate is $\frac12\cdot\frac12\cdot\frac12\cdot9! = \frac18\cdot9\cdot8\cdot7! = 9\cdot7!$ which is answer choice $\mathrm{(A)}$.

Alternatively, note that we can choose ${9\choose 2}$ places for the 1 and 2, then ${7\choose2}$ places for the 3 and 4, then $5\choose 2$ places for the 5 and 6, and the arrange the 7, 8 and 9 in $3!$ ways, giving us a total of ${9\choose2}\cdot{7\choose2}\cdot{5\choose2}\cdot3! = \frac{(9\cdot8)\cdot(7\cdot6)\cdot(5\cdot4)\cdot3!}{2\cdot2\cdot2} = 9\cdot7! \Longrightarrow \mathrm{(A)}$.