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A drawer contains red socks, black socks, and white socks. What is the least number of socks that must randomly be taken out of the drawer to be sure of having four pairs of socks? (A pair is two socks of the same color.)

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09 Oct 2010, 10:09

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Other wrote:

A drawer contains red socks, black socks, and white socks. What is the least number of socks that must be taken out of the drawer to be sure of having 4 matching pairs of socks?

a ) 8 B ) 10 C ) 12 D ) 14 E ) 16

The answer is 10, hence option B.

We need to find 4 matching pair of socks of one color. Take the worst case scenario:

First we pull one red , then one black , then one white. same we repeat for 3 times. Now we have 9 socks but only 3 pairs of of socks. The next socks we pull could be of any of the color and this will make 4th pair.

Red (1, 1,) (1, 1) Black (1, 1,) 1 White (1,1),1

So total 10.
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The only way of finding the limits of the possible is by going beyond them into the impossible.

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09 Oct 2010, 11:28

how many of each type/color of socks are there? If there are 100 of each lets say... then it could so happen that you pull out 100 black socks one after the other n your first 100 tries... how can there be a guarantee of being able to draw 4 pairs unless we know how many of each color are present?

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09 Oct 2010, 12:39

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gmat1011 wrote:

how many of each type/color of socks are there? If there are 100 of each lets say... then it could so happen that you pull out 100 black socks one after the other n your first 100 tries... how can there be a guarantee of being able to draw 4 pairs unless we know how many of each color are present?

Question asks 'what is least possible number' so we have to take the above mentioned scenario.
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The only way of finding the limits of the possible is by going beyond them into the impossible.

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09 Oct 2010, 19:41

it is the 'least number' but it is in the same sentence tied to "to be sure" of taking out... if you draw 10 randomly you cannot be sure that you will get 4 pairs - you cannot guarantee that result...

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10 Oct 2010, 00:07

gmat1011 wrote:

how many of each type/color of socks are there? If there are 100 of each lets say... then it could so happen that you pull out 100 black socks one after the other n your first 100 tries... how can there be a guarantee of being able to draw 4 pairs unless we know how many of each color are present?

If you pull 100 black socks one after the other, you'd have 4 pairs after eight draws.

gmat1011 wrote:

it is the 'least number' but it is in the same sentence tied to "to be sure" of taking out... if you draw 10 randomly you cannot be sure that you will get 4 pairs - you cannot guarantee that result...

Yes you would be, thats what has been explained above Just think of how many you can keep taking out while ensuring no pair is formed. R, B, W 4th draw, you will form a pair for sure R 5th you can avoid R 6th draw will always be a pair in that case B 7th you aviod B 8th will be a pair W 9th you avoid W 10th draw has to be a pair R
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10 Oct 2010, 04:29

but what if there are 30 socks of each color - the question doesn't mention how many there are of each. then worse case in your first 10 draws you may pick 10 blacks socks in a row without forming a pair. how can you be sure of forming pairs?

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10 Oct 2010, 11:52

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gmat1011 wrote:

but what if there are 30 socks of each color - the question doesn't mention how many there are of each. then worse case in your first 10 draws you may pick 10 blacks socks in a row without forming a pair. how can you be sure of forming pairs?

I think you are interpreting this question incorrectly. As soon as you draw 2 black socks you have a pair. So if there are only black socks, with every two you keep forming a pair and you will be done in just 8 socks.
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12 Oct 2010, 10:49

Even I got but not convienced much since I hv consider that first 3 picks -3 diff colors...then onwards I solved it to get the and 10 but y should I assume as my first 3draws of diff colors??

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12 Oct 2010, 11:31

vitamingmat wrote:

Even I got but not convienced much since I hv consider that first 3 picks -3 diff colors...then onwards I solved it to get the and 10 but y should I assume as my first 3draws of diff colors??

Posted from my mobile device

Its a "greedy" way of solving the problem. You need to find the maximum number of ways, in order to do this you pick socks in a way such that it takes the maximum number of picks to form pairs. In order to do this you delay forming pairs as much as you can. If you choose any other way of picking the first three socks (all same OR two same one different) after three picks you already have a pair, but if you choose 3 different socks, you have no pairs.
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A drawer contains red socks, black socks, and white socks. What is the least number of socks that must randomly be taken out of the drawer to be sure of having four pairs of socks? (A pair is two socks of the same color.)

A. 8 B. 10 C. 12 D. 14 E. 16

A good Q, which becomes easy the moment we understand the logic behind it... 4 pair of socks means there will be atleast two pairs of one colour..

since the Q uses words"to be sure", we take the worst case where we pick a different colour each time we pick..

the worst case would be we have 3 of each colour, such that the one picked up after that fulfills the condition of four pair...

3 of each colour means 1 pair +1 of each colour.. so total =3*3=9.. now the tenth we pick, irrespective of the colour picked, would make it four pairs.. so total =9+1=10 .. B
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