Arrangements of numbers-odd and even positions

  • Context: MHB 
  • Thread starter Thread starter evinda
  • Start date Start date
  • Tags Tags
    even
Click For Summary

Discussion Overview

The discussion revolves around a combinatorial exercise involving the arrangements of the numbers from 1 to 2n, specifically focusing on the placement of even and odd numbers in even and odd positions. Participants explore how many arrangements can be made under certain conditions, including the requirement that even positions contain only even numbers and the presence of at least one even number in any position.

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • Some participants propose that there are $n! \cdot n!$ arrangements where even positions contain only even numbers.
  • Others suggest that the number of arrangements with at least one even number might be calculated as $\binom{n}{1} \cdot (n-1)! \cdot n!$ or as $(2n)! - n! \cdot n!$.
  • A participant questions the problem statement, suggesting that it may imply a requirement for at least one even number to be in an even position, leading to a total of $(2n)!$ arrangements.
  • There is acknowledgment of a potential typo in the problem statement regarding the conditions for even numbers in positions.

Areas of Agreement / Disagreement

Participants express differing views on the interpretation of the problem and the calculations involved. There is no consensus on the correct approach or final answer, and some participants believe the problem is resolved while others remain uncertain.

Contextual Notes

There are unresolved assumptions regarding the interpretation of the problem statement and the conditions for counting arrangements. The discussion reflects varying interpretations of the requirements for even and odd placements.

evinda
Gold Member
MHB
Messages
3,741
Reaction score
0
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? :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? (Blush)
 
Physics news on Phys.org
evinda said:
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? :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? (Blush)

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

evinda said:
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. :eek:

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)
 
I like Serena said:
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. :eek:

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

Okay. Then it's all SOLVED I guess! (Whew)
 
I like Serena said:
Okay. Then it's all SOLVED I guess! (Whew)

Yes! (Party)
 

Similar threads

Undergrad What is the probability that the product of the digits is odd?
  • · Replies 13 ·
Replies
13
Views
3K
MHB Find the smallest positive integer N
  • · Replies 3 ·
Replies
3
Views
1K
If p is even and q is odd, then ##\binom{p}{q}## is even
  • · Replies 19 ·
Replies
19
Views
3K
MHB Where Did the Monte Carlo Simulation Go Wrong?
  • · Replies 1 ·
Replies
1
Views
2K
Undergrad Determining the number of arrangements (complicated)
  • · Replies 7 ·
Replies
7
Views
2K
MHB Probability of misprints on each page
  • · Replies 7 ·
Replies
7
Views
2K
MHB What is the Minimum Number of Friends Needed for Unique Dinner Invitations?
  • · Replies 1 ·
Replies
1
Views
1K
MHB Find the probability that each group has an equal amount of odd and even numbers
  • · Replies 1 ·
Replies
1
Views
2K
MHB Solving Permutations Question with Restrictions
  • · Replies 3 ·
Replies
3
Views
1K
Undergrad Ways to put n balls in m boxes such that exactly k boxes are empty?
  • · Replies 29 ·
Replies
29
Views
5K