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An insect has one shoe and one sock for each of its twelve [#permalink]

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02 Dec 2009, 00:35

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61% (01:30) wrong based on 70 sessions

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An insect has one shoe and one sock for each of its twelve legs. In how many different orders can the insect put on its socks and shoes, assuming that, on each leg, the sock must be put on before the shoe?

Also I believe nothing like this will ever occur at real test, as this question is beyond the scope of GMAT.

Anyway here is my solution:

NOTE that each sock and shoe is "assigned" to a specific leg.

Imagine situation with no restriction, meaning no need to put the socks before the shoes. In this case the # of ways insect can put 24 items would be 24!. As we can choose to put ANY of 24 items first, then 23 items left, then 22 and so on.

Next step. On EACH leg we can put either sock OR shoe first. But for EACH leg from 12, only one order is correct WITH restriction: sock first then shoe. For one leg chances of correct order is 1/2, for two legs 1/2^2, similarly for 12 legs chances of correct order would 1/2^12.

So we get that for the total # of ways, WITH NO RESTRICTION, which is 24!, only 1/2^12 is good WITH RESTRICTION.

Trying to fully understand Bunuel's method. When we arrange them all in 24! ways, leg could possibly have 1 shoe + 1 sock, 1 sock + 1 shoe, 2 socks or 2 shoes.

Do we not need to remove out the cases where the leg has only shoes or only socks before imposing the restriction and dividing by 2?

Trying to fully understand Bunuel's method. When we arrange them all in 24! ways, leg could possibly have 1 shoe + 1 sock, 1 sock + 1 shoe, 2 socks or 2 shoes.

Do we not need to remove out the cases where the leg has only shoes or only socks before imposing the restriction and dividing by 2?

Leg can not have two socks or two shoes, as each leg has its specific shoe or specific sock. Meaning that shoe and sock is fixed to a certain leg, hence insect can put on one leg ONLY the sock and shoe of THIS leg.

my goodness...this is some heavy question...and i thought i was good at combinations! bunuel's method of combining combinatorics and probability is sheer genius. +1!
_________________

It matters not how strait the gate, How charged with punishments the scroll, I am the master of my fate : I am the captain of my soul. ~ William Ernest Henley

Next step. On EACH leg we can put either sock OR shoe first. But for EACH leg from 12, only one order is correct WITH restriction: sock first then shoe. For one leg chances of correct order is 1/2, for two legs 1/2^2, similarly for 12 legs chances of correct order would 1/2^12.

Hi bunel, can you please explain the 1/2 part may be with a simpler example.I'm dumb

Also I believe nothing like this will ever occur at real test, as this question is beyond the scope of GMAT.

Anyway here is my solution:

NOTE that each sock and shoe is "assigned" to a specific leg.

Imagine situation with no restriction, meaning no need to put the socks before the shoes. In this case the # of ways insect can put 24 items would be 24!. As we can choose to put ANY of 24 items first, then 23 items left, then 22 and so on.

Next step. On EACH leg we can put either sock OR shoe first. But for EACH leg from 12, only one order is correct WITH restriction: sock first then shoe. For one leg chances of correct order is 1/2, for two legs 1/2^2, similarly for 12 legs chances of correct order would 1/2^12.

So we get that for the total # of ways, WITH NO RESTRICTION, which is 24!, only 1/2^12 is good WITH RESTRICTION.

So the final answer is 24!/2^12.

Answer: C.

Bunuel, Sorry to open this thread. Why wouldn't it be 12! * 12!? For each of the legs, there are 12 possible socks = 12!. Once the socks are chosen, there are 12 possible shoes = 12!

Therefore, using Fund. Principle of counting, the total permutations should be (12!)^2 ? Correct?

Also I believe nothing like this will ever occur at real test, as this question is beyond the scope of GMAT.

Anyway here is my solution:

NOTE that each sock and shoe is "assigned" to a specific leg.

Imagine situation with no restriction, meaning no need to put the socks before the shoes. In this case the # of ways insect can put 24 items would be 24!. As we can choose to put ANY of 24 items first, then 23 items left, then 22 and so on.

Next step. On EACH leg we can put either sock OR shoe first. But for EACH leg from 12, only one order is correct WITH restriction: sock first then shoe. For one leg chances of correct order is 1/2, for two legs 1/2^2, similarly for 12 legs chances of correct order would 1/2^12.

So we get that for the total # of ways, WITH NO RESTRICTION, which is 24!, only 1/2^12 is good WITH RESTRICTION.

So the final answer is 24!/2^12.

Answer: C.

Bunuel, Sorry to open this thread. Why wouldn't it be 12! * 12!? For each of the legs, there are 12 possible socks = 12!. Once the socks are chosen, there are 12 possible shoes = 12!

Therefore, using Fund. Principle of counting, the total permutations should be (12!)^2 ? Correct?

hi voodoochild, For each of the legs, there are possible 24 objects (12 shoes+12 socks) that can be worn. therefore 24! With the restriction given, in half of the orders, a shoe would be worn before sock and in the other half, a sock would be worn before the shoe . C seems to be the only option with that restriction. IMO 24!/2 would be the answer. I dont understand the need to restrict the order i.e (1/2)^12 times.

Re: An insect has one shoe and one sock for each of its twelve [#permalink]

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05 Aug 2012, 01:18

Bunuel wrote:

OA for this question is C.

BUT: this is question is out of the scope of GMAT, so I wouldn't worry about it at all. No need to waste much of your time on it.

Hi Bunuel,

I think this is an advanced version of a question I saw on the forum before... something about people standing in a queue and it was about X should stand in front of Y (there were specific names). There were n (some specific number) people altogether, so the number of possibilities was n!/2 (in half of them X was in front of Y and in the other half X wasn't). Can you pull out the question? The insect question would be a version of the queue question, for example with 24 couples (husbands and wives), and asking for the number of possibilities in which each wife stands in front of her husband. Or in general: total of \(n\) people in a queue, \(k\) couples (\(2k \leq n\)), each wife standing in front of her husband - number of possible arrangements is \(n!/(2^k)\).

It could be a +700 question on a test...don't you think?
_________________

PhD in Applied Mathematics Love GMAT Quant questions and running.

BUT: this is question is out of the scope of GMAT, so I wouldn't worry about it at all. No need to waste much of your time on it.

Hi Bunuel,

I think this is an advanced version of a question I saw on the forum before... something about people standing in a queue and it was about X should stand in front of Y (there were specific names). There were n (some specific number) people altogether, so the number of possibilities was n!/2 (in half of them X was in front of Y and in the other half X wasn't). Can you pull out the question? The insect question would be a version of the queue question, for example with 24 couples (husbands and wives), and asking for the number of possibilities in which each wife stands in front of her husband. Or in general: total of \(n\) people in a queue, \(k\) couples (\(2k \leq n\)), each wife standing in front of her husband - number of possible arrangements is \(n!/(2^k)\).

It could be a +700 question on a test...don't you think?

GMAT combination/probability questions are fairly straightforward, so frankly speaking I think that there is 0 chances for this kind of question appearing on the real test.

BUT: this question is out of the scope of GMAT, so I wouldn't worry about it at all. No need to waste much of your time on it.

Bunuel, this is tough indeed. Just for the peace of mind, I thought of this question as Leg 1 has 12 socks and 12 shoes, Leg 2 has 11 socks and 11 shoes...etc. Again, it feels a bit more intuitive...what do you think of this approach? I am struggling to put it into the answers and come up with an algebraic way to get to the answer...Would not it be \((12*12)+(11*11)+(10*10)...(1*1)\)? Please share your thoughts...

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