Question

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

Answer 1

@oldrin.bataku @mathslover

Answer 2

http://www.wolframalpha.com/input/?i=derangement+5

Answer 3

haha,i was thinking of applying the dearrangement formula itself!

Answer 4

but the answer 44 is incorrect :|

Answer 5

you mean they can sit arbitrarily?

Answer 6

maybe

Answer 7

what we can do is first seat all the women in 10C5 ways..to start the question :D

Answer 8

interesting

Answer 9

haha didn't you notice all my questions are interesting:P

Answer 10

possibly could be
http://www.wolframalpha.com/input/?i=Sum%5BBinomial%5B5%2Cn%5D%5E2%2C+%7Bn%2C+0%2C+5%7D%5D * 44

Answer 11

not in options..

Answer 12

is C) correct answer?

Answer 13

haha no :|
bad luck today..

Answer 14

now the answer is obvious

Answer 15

lol

Answer 16

guessing as well as combinatorics is not my best part. still I can't think way how.

Answer 17

@oldrin.bataku has something big to say :O

Answer 18

i hope so.

Answer 19

oh u asked this same question 7 months ago?

Answer 20

lol yeah,no solutions obtained that time either

Answer 21

looks like i missed few things, what if both husband and wives are in same row.

Answer 22

anyone can refer the older posts on the same question for some ideas
http://openstudy.com/study#/updates/514ea923e4b0ae0b658b15e0

Answer 23

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 \(30240\times44=1330560\)

Answer 24

(b) is your answer and the form is trivial to reach. Recognize \(_{10}P_5=_{10}C_5\times5!=_{10}C_5\times120\).

Answer 25

10 rows?what do u mean O_O

Answer 26

oops 10 seats

Answer 27

isn't it possible that 2 wives sit behind each other?

Answer 28

what if a man has 2 wives

Answer 29

Tsk tsk.

Answer 30

:-p

Answer 31

@DLS as it turns out it makes no difference :-p

Answer 32

okay,i have one last,yet interesting question on this topic!

Answer 33

@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 \(5!=44\).