Total number of permutations of 1 to 10 such that for $i=1,2,3,4,5$ we have $2i-1$ before $2i$.

Question

We have ten numbers $\{1,2,3,...,10\}$. We want the total number of permutations of these ten numbers in which for every $i\in\{1,2,3,4,5\}$, there appears $2i-1$ before $2i$. Appreciated any help in advance.

Answer 1

10! / (2^5)

supposed you have: .... 2i-1 .... 2i ...

for every such permutation you have the mirror one which is .... 2i .... 2i-1 ... which means you will have to take only half of those permutations. Hense you have 5 such conditions you will have to divide to 2*2*2*2*2 and all the permutations from 1 to 10 are 10!

Answer 2

Suppose we have $2n+2$ numbers and let $a_{n+1}$ be the number of good permutations. Then however we choose a two element subset from $\{1,2,...,2n+1,2n+2\}$ we put on a smaller number $2n+1$ and a bigger $2n+2$. So we have to arrange the rest of $2n$ numbers and that we can do on $a_n$ good ways. Thus we have $$a_{n+1} = {2n+2\choose 2}a_n$$ where $a_1 =1$.

Thus $$a_5 = {10\choose 2}\cdot {8\choose 2}\cdot {6\choose 2}\cdot {4\choose 2}\cdot 1 = {10!\over 2^5}$$