4 chairs are there : GMAT Problem Solving (PS)

4 chairs are there

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4 chairs are there [#permalink]

26 Nov 2009, 09:11

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4 chairs are there. Two person A and B wants to sit there such that A is always left to B. How many ways it can be done ?

Can we write it as 4C2 way ? I am doubtful as there is a constraint " A is always left to B".

whats your comments ?

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Re: 4 chairs are there [#permalink]

26 Nov 2009, 09:47

There a few ways how to approach this problem:

1) Enumeration of possibilities

ABxx
AxBx
AxxB
xABx
xAxB
xxAB

Answer: 6

2)
A at 1st left place: C^3_1 = 3 ways for B
A at 2st left place: C^2_1 = 2 ways for B
A at 3st left place: C^1_1 = 1 ways for B

Answer: 6

3) trick: symmetry

P^4_2 = 12 - the total number of ways
A is always left to B in exactly 1/2 of all cases.

Answer: 1/2 * P^4_2 = 6
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Re: 4 chairs are there [#permalink]

26 Nov 2009, 10:11

instead of doing your way , can i just use the one liner combination rule ?

i.e pick 2 chairs out of 4 ==> 4C2 = 6

As you see I also got the same answer as you got ....i just applied the combination definition rule

But is it a valid math ? That was intent of this question .
If its not valid math , then explain why its not

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Re: 4 chairs are there [#permalink]

26 Nov 2009, 10:23

if you apply 4C2, you need to understand why. 4C2 means how many ways you have to choose 2 chairs out of 4 chairs but it is not enough.

4C2 is right if you think this way:

1) let's choose any 2 seats out of 4.
2) For chosen 2 seats we pick left seat for A and right seat for B.
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Re: 4 chairs are there [#permalink]

26 Nov 2009, 13:14

walker wrote:

3) trick: symmetry

P^4_2 = 12 - the total number of ways
A is always left to B in exactly 1/2 of all cases.

Answer: 1/2 * P^4_2 = 6

I think this is the best way.

Variation without P and combinatorial formulas:

"A" can pick any of 4 chairs. In each case there would be 3 other chairs "B" could sit on.

4*3=12 - # of total possibilities. 12/2=6 - # of ways they sit when "A" is left to "B"

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Re: 4 chairs are there [#permalink]

26 Nov 2009, 19:25

walker wrote:

but that red colored constraint is a violation on the very definition of combination ....definition of combination talks about group ...it does not bother about specific order....is not it ? ....so, we really cant write down 4C2 here as per the combination rule .

I am happy with your earlier solution.

I found in a book ...they used 4C2 in the solution . I object this ...because this is against the definition of combination ( as there is a constraint).

But I observe you did not use this 4C2 thingies and hence I liked your solution.

comments please.

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Re: 4 chairs are there [#permalink]

26 Nov 2009, 22:54

4C2 means the number of ways to choose 2 objects out of 4. No more, no less. So, nothing is wrong with using 4C2 in my approach.
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Re: 4 chairs are there [#permalink]

28 Nov 2009, 00:51

walker wrote:

4C2 means the number of ways to choose 2 objects out of 4. No more, no less. So, nothing is wrong with using 4C2 in my approach.

what if the question asked was

4 chairs are there. Two person A and B wants to sit there . How many ways it can be done ?

//note i removed the clause "such that A is always left to B".

is it still the same answer 4C2 ? does that clause have no impact at all ? why ?

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Re: 4 chairs are there [#permalink]

28 Nov 2009, 07:06

There are four places and two of them want to sit in only one way. If we combine two (A & B) places into one, then there are 6 ways to occupy three places and 1 way to arrange the A and B.

The total ways is 6*1=6.

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Re: 4 chairs are there [#permalink]

28 Nov 2009, 07:07

ghentu wrote:

walker wrote:

4C2 means the number of ways to choose 2 objects out of 4. No more, no less. So, nothing is wrong with using 4C2 in my approach.

Then the answer would be 6*2=12 ways

This is an arrangement and not a selection. IMO there is no need to use Combination formula

Re: 4 chairs are there [#permalink] 28 Nov 2009, 07:07

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4 chairs are there

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