DLS
  • DLS
There are two rows ,one behind other, of 5 chairs each . Five couples are to be seated , number of arrangements such that no husband sits in front of or behind his wife (a) 120x44 (b) 10C5x120x44 C)10C5 x 44(d) 44
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
DLS
  • DLS
@oldrin.bataku @mathslover
experimentX
  • experimentX
http://www.wolframalpha.com/input/?i=derangement+5
DLS
  • DLS
haha,i was thinking of applying the dearrangement formula itself!

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DLS
  • DLS
but the answer 44 is incorrect :|
experimentX
  • experimentX
you mean they can sit arbitrarily?
DLS
  • DLS
maybe
DLS
  • DLS
what we can do is first seat all the women in 10C5 ways..to start the question :D
dan815
  • dan815
interesting
DLS
  • DLS
haha didn't you notice all my questions are interesting:P
experimentX
  • experimentX
possibly could be http://www.wolframalpha.com/input/?i=Sum%5BBinomial%5B5%2Cn%5D%5E2%2C+%7Bn%2C+0%2C+5%7D%5D * 44
DLS
  • DLS
not in options..
experimentX
  • experimentX
is C) correct answer?
DLS
  • DLS
haha no :| bad luck today..
DLS
  • DLS
now the answer is obvious
dan815
  • dan815
lol
experimentX
  • experimentX
guessing as well as combinatorics is not my best part. still I can't think way how.
DLS
  • DLS
@oldrin.bataku has something big to say :O
experimentX
  • experimentX
i hope so.
dan815
  • dan815
oh u asked this same question 7 months ago?
DLS
  • DLS
lol yeah,no solutions obtained that time either
experimentX
  • experimentX
looks like i missed few things, what if both husband and wives are in same row.
DLS
  • DLS
anyone can refer the older posts on the same question for some ideas http://openstudy.com/study#/updates/514ea923e4b0ae0b658b15e0
anonymous
  • anonymous
Consider any single seating of the 5 women in the 10 rows. You want to consider then their spouses arranged in the seats right in front of or in back of them, and wish to arrange them such that they don't sit in the same column as their spouse; considering numbered couples, for example, given 1 2 3 4 5 for the women, we'd want 2 3 4 5 1, for example, or 5 1 2 3 4, etc., for the men. This is the number of derangements of length 5, given by \(!5=44\). Now how many possible seatings of the 5 women are there? Simple: \(_{10}P_5=10!/5!=30240\). Our total number of seats is then just given by \(30240\times44=1330560\)
anonymous
  • anonymous
(b) is your answer and the form is trivial to reach. Recognize \(_{10}P_5=_{10}C_5\times5!=_{10}C_5\times120\).
DLS
  • DLS
10 rows?what do u mean O_O
anonymous
  • anonymous
oops 10 seats
DLS
  • DLS
isn't it possible that 2 wives sit behind each other?
dan815
  • dan815
what if a man has 2 wives
Jhannybean
  • Jhannybean
Tsk tsk.
anonymous
  • anonymous
:-p
anonymous
  • anonymous
@DLS as it turns out it makes no difference :-p
DLS
  • DLS
okay,i have one last,yet interesting question on this topic!
anonymous
  • anonymous
@DLS all that matters is that you can assign a couple number to each person in each row and make it so the rows have different derangements of couple numbers... 5 couples yields \(5!=44\).

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