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# Six distinct pair of shoes are in a closet. If three shoes are

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Six distinct pair of shoes are in a closet. If three shoes are  [#permalink]

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Updated on: 23 Apr 2017, 01:57
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Question Stats:

61% (02:18) correct 39% (02:37) wrong based on 177 sessions

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Six distinct pair of shoes are in a closet. If three shoes are selected at random from the closet,what is the probability that no two of the chosen shoes make a pair

A. 1/5
B. 2/11
C. 13/33
D. 8/11
E. 4/5

Originally posted by VyshakhR1995 on 22 Apr 2017, 23:29.
Last edited by Bunuel on 23 Apr 2017, 01:57, edited 2 times in total.
Renamed the topic and edited the question.
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Re: Six distinct pair of shoes are in a closet  [#permalink]

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23 Apr 2017, 01:58
1
4
VyshakhR1995 wrote:
Six distinct pair of shoes are in a closet.if three shoes are selected at random from the closet,what is the probability that no two of the chosen shoes make a pair

1/5
2/11
13/33
8/11
4/5

Hi,
Please post Q in correct format with choices marked as A,B....

There are 6distinct pairs, so 12 shoes in all...
Way to choose 3 out of 12 =12*11*10..
Way to choose all 3 from different pairs:-
1) first can be any of 12.....So 12 ways
2) second can be any of remaining except the pair of 1st choosen, do 11-1=10 ways..
3) third can be any of remaining 10, except of pairs of first two..., So 10-2*1=8..

Probability=$$\frac{12*10*8}{12*11*10}=\frac{8}{11}$$
D
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Re: Six distinct pair of shoes are in a closet. If three shoes are  [#permalink]

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12 Sep 2017, 03:35
3
1
VyshakhR1995 wrote:
Six distinct pair of shoes are in a closet. If three shoes are selected at random from the closet,what is the probability that no two of the chosen shoes make a pair

A. 1/5
B. 2/11
C. 13/33
D. 8/11
E. 4/5

I use method 1 - opposite probability

Select 3 shoes with 2 are a pair:

Ways to select a pair: 6C1
Ways to select any shoe from 10 shoes: 10C1
Total ways to choose 3 shoes: 12C3

Probability: (6C1X10C1)/12C3 = 3/11

Probability of no pair of shoes in any three shoes: 1-(3/11)=8/11

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Re: Six distinct pair of shoes are in a closet. If three shoes are  [#permalink]

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19 Jul 2017, 15:00
2
There are 12 shoes, in 6 pairs. It doesn't matter what shoe we pick first. There will then be 11 shoes left, and 10 of those don't match our first shoe. So the probability our second selection doesn't make a pair is 10/11. Then we pick a third shoe from the 10 that remain, and there are 8 shoes left that won't make a pair (we need to pick a shoe from one of the four unused pairs of shoes that remain). We need both those things to happen, so we multiply the two probabilities: the answer is (10/11)*(8/10) = 8/11
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Six distinct pair of shoes are in a closet. If three shoes are  [#permalink]

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22 Jun 2017, 14:06
1
VyshakhR1995 wrote:
probably a very basic question but this is confusing me:

When I think of choosing 3 items from 12. I think of the number of ways = 12C3 which is not 12*11*10 rather 220. What am I doing wrong.

Didn't get your question.Can you explain??

12C3

$$\frac{12*11*10}{3*2*1}$$=220

This is what is meant. Why don't you divide by 6?
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Re: Six distinct pair of shoes are in a closet. If three shoes are  [#permalink]

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12 Sep 2017, 02:08
1
1
sasidharrs wrote:
Hi,
1st selection probability is 1 (choose anyone)
2nd selection probability is 1/10( since we should not select same pair of shoe)
3rd selection probability is 1/8(since we should not select the first 2 pair color shoes)
total probability : 1*1/10*1/8 = 1/80

But, my answer is wrong, I do not know why.. basically, I do am not able to decide when to use combination approach?

Your approach is almost there - the bit you are missing is thinking about the question as odds it is NOT a pair.

1st selection probability is correctly identified as 1
2nd selection probability is (12-1=11 shoes left, of which 1 is the same pair): 10/11
3rd selcetion probability is (11-1=10 shoes left, of which 2 are the same pair): 8/10

8/10*10/11 = 80/110 = 8/11

Thinking & maths was right, but problem approach was the wrong way around! Hope that helped.
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Re: Six distinct pair of shoes are in a closet. If three shoes are  [#permalink]

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21 Jun 2017, 19:44
probably a very basic question but this is confusing me:

When I think of choosing 3 items from 12. I think of the number of ways = 12C3 which is not 12*11*10 rather 220. What am I doing wrong.
Senior Manager
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Posts: 252
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Re: Six distinct pair of shoes are in a closet. If three shoes are  [#permalink]

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22 Jun 2017, 11:28
probably a very basic question but this is confusing me:

When I think of choosing 3 items from 12. I think of the number of ways = 12C3 which is not 12*11*10 rather 220. What am I doing wrong.

Didn't get your question.Can you explain??
Senior Manager
Joined: 02 Mar 2017
Posts: 252
Location: India
Concentration: Finance, Marketing
Re: Six distinct pair of shoes are in a closet. If three shoes are  [#permalink]

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24 Jun 2017, 21:01
Smokeybear00 wrote:
VyshakhR1995 wrote:
probably a very basic question but this is confusing me:

When I think of choosing 3 items from 12. I think of the number of ways = 12C3 which is not 12*11*10 rather 220. What am I doing wrong.

Didn't get your question.Can you explain??

12C3

$$\frac{12*11*10}{3*2*1}$$=220

This is what is meant. Why don't you divide by 6?

Now i think i understand the original question,but now i am confused by your's.
12*11*10 gives the arrangement.
If A B C are three items when considering arrangement we account for ACB BCA etc
For Combination we Don't need to worry about that.
If you are selecting a committee,it doesn't matter whether you Choose A or B first.But when you are arranging them you need to consider the order.
The formula you mentioned calculates the arrangement.
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Re: Six distinct pair of shoes are in a closet. If three shoes are  [#permalink]

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19 Jul 2017, 14:56
This question needs a detailed explanation involving multiple approaches. Can somebody help here?
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Re: Six distinct pair of shoes are in a closet. If three shoes are  [#permalink]

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22 Jul 2017, 13:50
Hi,
1st selection probability is 1 (choose anyone)
2nd selection probability is 1/10( since we should not select same pair of shoe)
3rd selection probability is 1/8(since we should not select the first 2 pair color shoes)
total probability : 1*1/10*1/8 = 1/80

But, my answer is wrong, I do not know why.. basically, I do am not able to decide when to use combination approach?
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Joined: 07 Nov 2016
Posts: 53
GMAT 1: 760 Q51 V42
Re: Six distinct pair of shoes are in a closet. If three shoes are  [#permalink]

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02 Jan 2019, 21:42
probably a very basic question but this is confusing me:

When I think of choosing 3 items from 12. I think of the number of ways = 12C3 which is not 12*11*10 rather 220. What am I doing wrong.

Total number of ways of selecting 3 items from 12 is indeed $$12C3$$= 220.

This is the denominator while calculating Probability

To find the number of ways in which no two shoes are from the same pair

First Shoe : 12 ways (any of the 12 shoes)
Second Shoe : 10 ways (any of the 10 shoes remaining after excluding the shoe remaining from the first pair)
Third Shoe : 8 ways

Total = 12 * 10 * 8 ways = 960

But this gives us the number of arrangements (where order matters)

To get the number of selections , we divide 960 by 3! to get 160

Required Probability = $$\frac{160}{220}$$ = 8/11
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Re: Six distinct pair of shoes are in a closet. If three shoes are  [#permalink]

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18 Jan 2020, 02:19
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Re: Six distinct pair of shoes are in a closet. If three shoes are   [#permalink] 18 Jan 2020, 02:19
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