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4 married couple have bought 8 seats in a raw for a football [#permalink]
20 Jul 2004, 08:24

00:00

A

B

C

D

E

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct
0% (00:00) wrong based on 0 sessions

4 married couple have bought 8 seats in a raw for a football game.

b) In how many ways can they be seated if each couple is not to sit together with the husband to the left of his wife?
c) In how many ways can they be seated if each couple is not to sit together?
d) In how many ways can they be seated if all men are to sit together?
e) In how many ways can they be seated if none of the men are to sit together and none of the women are sit together?

b) In how many ways can they be seated if each couple is not to sit together with the husband to the left of his wife?

total number of ways 4 couples can sit is 8!
number of ways couples can sit together with husbands sitting to the left of wife : since the wife husband cant be split and have to sit together it is 4! ways

whwhwhwh

so ans to b is 8! - 4!

C) similarly couples can sit together in 4!*2 ways
(whwhwhwh and hwhwhwhw)

so the ans is 8! - 4!*2 ways

the rest of the answers remain the same
D) 5!*4! = 2280
E) 2*4!*4! = 1152

Boksana, just saw Ur answers, but still thot of posting my answers... btw, any explanation for your answers? Ur answers seems too so small numbers though for a 8!...

b) In how many ways can they be seated if each couple is not to sit together with the husband to the left of his wife?
Total number of ways they can be seated = 8! Number of ways they can be seated - with each couple sitting together w/ h left to w = 4! (within a couple of HW is fixed and need not be played with) So, ans for b = 8! - 4!

c) In how many ways can they be seated if each couple is not to sit together?
Total number of ways they can be seated = 8! Number of ways they can be seated - with each couple sitting together = 4!*2! (within a couple of HW and WH can be played with) So, ans for c = 8! - 4!*2!

d) In how many ways can they be seated if all men are to sit together?
Number of entities to seat = 5, hence ways = 5! Number of ways men can be seated among themselves = 4! SO, anc for d = 5!*4!

e) In how many ways can they be seated if none of the men are to sit together and none of the women are sit together?
Total number of ways they can be seated = 8! Number of ways they can be seated - men sitting together and women sitting together = 2!*4!*4! (first 2! for two entities, the two 4! are for men and women to reorder within themselves) So, ans for d = 8! - 2!*4!*4!

[quote="hardworker_indian"]Boksana, just saw Ur answers, but still thot of posting my answers... btw, any explanation for your answers? Ur answers seems too so small numbers though for a 8!...

e) In how many ways can they be seated if none of the men are to sit together and none of the women are sit together?
[color=blue]Total number of ways they can be seated = 8!
Number of ways they can be seated - men sitting together and women sitting together = 2!*4!*4! (first 2! for two entities, the two 4! are for men and women to reorder within themselves)
So, ans for d = 8! - 2!*4!*4![/color][/quote]ct

hi hardworker_indian,

I'm not sure why u want to subtract 2!*4!*4! from 8!. remember that will give you all possible combinations where husband wife are together
whwhwhwh etc etc where as what we are looking for is wwww hhhh type of combinations (you have to imagine that there are spectators sitting in between them) and so the ans I think is 2!*4!*4!

What I found was wrong; my formula was
Number of ways, none of the H sit together and none of W sit together
=
total number of ways -
number of ways that H sit together and W sit together (HHHHWWWW and WWWWHHHH)
=
8! -
2!*4!*4!

The loophole was that this includes other cases like three H sitting together, which is not desired... (Eg: HHHWWWWH)