How many arrangements of the numbers $1,2,3, \dots, 2n-1,2n$ exist such that at the even positions there are only even numbers?
How many arrangements are there,such that at least at one even position there is an even number? :confused:
Since, at the set $\{1,2,3, \dots, 2n-1,2n \}$,there are $n$ even numbers and $n$ odd ones,are there $n! \cdot n!$ arrangements,such that at the even positions there are only even numbers,right?
At the second subquestion,is it maybe $\binom{n}{1} \cdot (n-1)! \cdot n!$ ,or am I wrong?
Best Answer
1) The even position are $n$, so there are $n! \cdot n! $ such permutations, because even numbers must stay in even positions and odd numbers in odd positions.
2) The answer is the total number of permutations minus the number of permutations in which all even numbers are in odd positions, so $$(2n)!- n!\cdot n!$$