MHB Arrangements of numbers-odd and even positions

[evinda]

Hello! :)

I am given the following exercise:
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 position there is an even number?

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? (Blush)

 

[evinda]

Hello! :)

I am given the following exercise:
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 position there is an even number?

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? (Blush)

At the second subquestion,it is $(2n)!-n! \cdot n!$,right?

 

[I like Serena]

Hello! :)

I am given the following exercise:
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 position there is an even number?

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? (Blush)

At the second subquestion,it is $(2n)!-n! \cdot n!$,right?

Hi! ;)

If I take the question literally, there will always be a position where there is an even number, so that would be $(2n)!$ arrangements.

It seems there is a typo in the problem statement though.
Can it be that the question should be that at least at one even position there should be an even number? (Wondering)

 

[evinda]

Hi! ;)

If I take the question literally, there will always be a position where there is an even number, so that would be $(2n)!$ arrangements.

It seems there is a typo in the problem statement though.
Can it be that the question should be that at least at one even position there should be an even number? (Wondering)

Yes,that's what I meant..I forgot a word... (Lipssealed)

 

[I like Serena]

Yes,that's what I meant..I forgot a word... (Lipssealed)

Okay. Then it's all SOLVED I guess! (Whew)

 

[evinda]

Okay. Then it's all SOLVED I guess! (Whew)

Yes! (Party)