Solution: 2017 Winter Final - 1
Author: Michiel SmidQuestion
Consider permutations $a_1,a_2,\dots,a_{10}$ of the set $\{1,2,\dots,10\}$ for which
- $a_1,a_3,a_5,a_7,a_9$ are all odd and
- $a_2,a_4,a_6,a_8,a_{10}$ are all even.
(a)
$10!$
(b)
$(5!)^2$
(c)
$5^5 \cdot 5^5$
(d)
$2 \cdot (5!)^2$
Solution
Well, let’s break it down
- We can place the first odd number in any of the 5 odd positions: 5
We can place the second odd number in any of the 4 remaining odd positions: 4
...
We can place the last odd number in the last remaining odd position: 1
This creates $ 5! $ permutations - We can place the first even number in any of the 5 even positions: 5
We can place the second even number in any of the 4 remaining even positions: 4
...
We can place the last even number in the last remaining even position: 1
This creates $ 5! $ permutations
Multiplying the two permutations, we get $ 5! \cdot 5! $