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Susan & John ready to fight as ever(combinatorics)

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by ashish1354 » Mon Sep 22, 2008 2:44 am
Find the numbers of ways in which 4 boys and 4 girls can be seated alternatively.

a)in a row and there is a boy named John and a girl named Susan amongst the group who cannot be put in adjacent seats
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Re: Susan & John ready to fight as ever(combinatorics)

by Morgoth » Mon Sep 22, 2008 3:06 am
ashish1354 wrote:Find the numbers of ways in which 4 boys and 4 girls can be seated alternatively.

a)in a row and there is a boy named John and a girl named Susan amongst the group who cannot be put in adjacent seats
IMO 980

OA?
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here's the OA

by ashish1354 » Mon Sep 22, 2008 3:25 am
648 is the original answer
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Re: here's the OA

by Morgoth » Mon Sep 22, 2008 3:52 am
ashish1354 wrote:648 is the original answer
Answer is indeed 648

BGBGBGBG = 4!*4! = 576

GBGBGBGB = 4!*4! = 576

total arrangements = 576 + 576 = 1152

To find the total arrangements lets take John & Susan together

JSBGBGBG = 3! * 3! * 7 = 252

We can arrange all the boys except for John in 3! ways
We can arrange all the girls except for Susan in 3! ways
We can arrange John & Susan together in 7 ways out of 8 places

Similarly,

SJGBGBGB = 3! * 3! * 7 = 252

252 + 252 = 504

1152 - 504 = 648
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by nil4700 » Mon Sep 22, 2008 3:57 am
They need 8 places in all.

- - - - - - - -

First keep John in first place, that leaves Sussane 3 places to be seated.

J G B G/S B G/S B G/S

so there are 3 ways. Now, the boys can be seated in 3! ways and girls also in 3! ways. ==> 3*3!*3!

But this arrangement is again possible if John is seated on the last seat also.

------- J

so => 2*3*3!*3! = 216 ways


now, if John is seated in the middle. Let's say 2 to 7 positions.

G J G B G/S B G/S B => 2*3!*3!

position 2,3,4,5 6 & 7 are 6*2*3!*3!

If position 4 or 6,



so ==> 6*2*3!*3! = 532

total = 648
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Re: here's the OA

by 4meonly » Mon Sep 22, 2008 8:43 am
Morgoth wrote: To find the total arrangements lets take John & Susan together
JSBGBGBG = 3! * 3! * 7 = 252

We can arrange all the boys except for John in 3! ways
We can arrange all the girls except for Susan in 3! ways
We can arrange John & Susan together in 7 ways out of 8 places
I generally agree with such aproach, but i have some doubts about such multiplying JSBGBGBG = 3! * 3! * 7 = 252
We can arrange John & Susan together in 7 ways out of 8 places
Multiplying by 7 assumes that such situation
BJSGBGBG is true, but as J is boy he cannot be the neighbour of any boy, we deal only with BG pairs.

What do you think? Are my doubts reasonable?
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by manulath » Fri Sep 26, 2008 8:51 am
let there be 8 places
1 2 3 4 5 6 7 8

Cases when John and Susan can be together

when boys are first, then John can occupy 1, 3, 5, 7
when he is on 1 Susan can occupy only 2 = 1possibility
when John on 3, Susan can take 2 and 4 = 2 Possibilities
likewise 2 possibilities each - when John takes 5 or 7
total possibilities = 1+2+2+2=7
rest 3 boys and girls can be arranged in 3!*3!
so it will be 7*3!*3!

likewise 7*3!*3! when girls come first

so cases when John and Susan SIt together = 2*7*3!*3! = 504

Total ways for arrangement for 4 boys and 4 girls = 2*4!*4! = 1152

cases when John and Susan dont sit together = 1152 - 504 = 648

------------------------------------------

[b]CAN SOMEONE EXPLAIN A SIMPLER AND FASTER WAY TO DO THIS - pls[/b]
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by ashish1354 » Wed Dec 16, 2009 10:26 pm
I have the same doubt as Meonly. When we calculate the number of combination with Susan and john together we still need to seat the boys and girls alternately because we are subtracting the combination from the total number of possibilities in which boys and girls are sitting alternately.

This is the reason i chose 4(number of ways the pair can sit together)x3!(number of ways girls arrange themselves)x3!(number of ways boys arrange themselves)x2(for the SJ gbgbgh and JSbgbgbg configurations)

can someone please tell me what is wrong with my approach.
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