Prob of choosing 6 socks so that there is no pair among them= 8C6 (2C1)^6
We are doing 2c1 coz any one one sock can b chosen among a pair
P (atleast one pair )= 1- p (no pair)

= 1- [ 8c6 * 2^6] / 16c6
=111/143

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Hi Brent,

I tried to use combinations. What am I doing wrong?

P(at least 1 pair) = 1 - P(no pairs)

P(no pairs) = (No. of ways 6 unpaired socks can be picked from 8 unpaired socks)/(total number of ways 6 socks can be picked from 16 socks)

= 8C6/16C6
= 1/286

P(at least 1 pair) = 1 - 1/286 = 285/286

Of course, answer not among choices.

Hi dramamur,

The problem with your solution is that you are treating the socks differently when calculating the numerator and denominator.
For the denominator, you are treating the 16 socks as being 100% unique. It's as though you are saying that each pair of matching socks has a LEFT sock and a RIGHT sock.
So, we can select 6 socks from 16 UNIQUE socks in 16C6 ways.

However, for the numerator, you are essentially saying that the two socks in each matching pair are identical.
You are not making a LEFT/RIGHT distinction for each pair. Instead, you are choosing 6 COLORS from a total of 8 COLORS.

Let's keep going with your solution, BUT we'll make sure we consider the fact that we have a LEFT and RIGHT sock for each pair.

So, we'll have to alter your numerator somewhat.
First, we can select 6 COLORS from a total of 8 COLORS in 6C2 ways (28 ways)
For the 1st COLOR, we can choose either the LEFT sock or the RIGHT sock. We can do this in 2 ways.
For the 2nd COLOR, we can choose either the LEFT sock or the RIGHT sock. We can do this in 2 ways.
For the 3rd COLOR, we can choose either the LEFT sock or the RIGHT sock. We can do this in 2 ways.
For the 4th COLOR, we can choose either the LEFT sock or the RIGHT sock. We can do this in 2 ways.
For the 5th COLOR, we can choose either the LEFT sock or the RIGHT sock. We can do this in 2 ways.
For the 6th COLOR, we can choose either the LEFT sock or the RIGHT sock. We can do this in 2 ways.
So, the total number of ways to select 6 UNMATCHED socks = (28)(2)(2)(2)(2)(2)(2)

We'll keep the numerator the same.
That is, we can select 6 socks from 16 UNIQUE socks in 16C6 ways.

So, P(ZERO matching socks) = (28)(2)(2)(2)(2)(2)(2) /16C6
= 32/143

P(at least 1 pair) = 1 - P(no pairs)
= 1 - 32/143
= 111/143

Cheers,
Brent
_________________

Brent Hanneson – GMATPrepNow.com

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