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Barbara has 8 shirts and 9 pants. How many clothing combinat

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Barbara has 8 shirts and 9 pants. How many clothing combinat  [#permalink]

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New post 03 Mar 2013, 10:05
2
9
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A
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Question Stats:

72% (01:12) correct 28% (01:39) wrong based on 184 sessions

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Barbara has 8 shirts and 9 pants. How many clothing combinations does Barbara have, if she doesn’t wear 2 specific shirts with 3 specific pants?

A. 41
B. 66
C. 36
D. 70
E. 56
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Re: Barbara has 8 shirts and 9 pants  [#permalink]

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New post 03 Mar 2013, 11:08
2
Total no of combinations available is= 8 shirts X 9 pants = 72
Eliminate the 2shirts X 3 pants combo = 6 which gives.......>> 72-6=66
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Re: Barbara has 8 shirts and 9 pants. How many clothing combinat  [#permalink]

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New post Updated on: 03 Apr 2013, 23:03
Clothing combination of shirt and pant left are 36 i.e
shirt=8-2=6
pant=9-3=6
so, shirt*pant, 6*6=36

Originally posted by michellark on 12 Mar 2013, 02:57.
Last edited by michellark on 03 Apr 2013, 23:03, edited 1 time in total.
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Re: Barbara has 8 shirts and 9 pants. How many clothing combinat  [#permalink]

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New post 12 Mar 2013, 03:21
8 shirts
9 pants

we can combine 2 shirts with (9-3) pants
2*6=12

we can combine the other shirts (6) with any pants (9)
6*9=54

Thus total :12+54=66
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Re: Barbara has 8 shirts and 9 pants. How many clothing combinat  [#permalink]

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New post 06 Aug 2013, 20:30
I have understood the above approaches . However, would like folks to answer that what is wrong with the below approach.

9 pants - 3 specific avoided - 6 pants
8 shirts - 2 specific avoided - 6 shirts

6C1 * 6C1 = 36 (select one from each for a combination)

What is wrong ?

Rgds,
TGC !
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Re: Barbara has 8 shirts and 9 pants. How many clothing combinat  [#permalink]

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New post 06 Aug 2013, 20:44
2
targetgmatchotu wrote:
I have understood the above approaches . However, would like folks to answer that what is wrong with the below approach.

9 pants - 3 specific avoided - 6 pants
8 shirts - 2 specific avoided - 6 shirts

6C1 * 6C1 = 36 (select one from each for a combination)

What is wrong ?

Rgds,
TGC !


hi TGC,

Question is: Barbara has 8 shirts and 9 pants. How many clothing combinations does Barbara have, if she doesn’t wear 2 specific shirts with 3 specific pants?

it means he is going to wear all pants and shirts except few combinations he is going to avoid.
so there are 2 groups now

6 shirts ==>they all have choices of 9 pants therefore total ways = \(6C1*9C1 = 54\)
NOW the left 2 shirts is going to avoid 3 pants therefore we have 6 pants hence total ways = \(2C1*6C1 =12\)

hence total choice = \(54 +12 =66\)

what you are doing is that you are just removing those shirts and pants.

hope it helps
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Re: Barbara has 8 shirts and 9 pants. How many clothing combinat  [#permalink]

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New post 07 Aug 2013, 00:03
blueseas wrote:
targetgmatchotu wrote:
I have understood the above approaches . However, would like folks to answer that what is wrong with the below approach.

9 pants - 3 specific avoided - 6 pants
8 shirts - 2 specific avoided - 6 shirts

6C1 * 6C1 = 36 (select one from each for a combination)

What is wrong ?

Rgds,
TGC !


hi TGC,

Question is: Barbara has 8 shirts and 9 pants. How many clothing combinations does Barbara have, if she doesn’t wear 2 specific shirts with 3 specific pants?

it means he is going to wear all pants and shirts except few combinations he is going to avoid.
so there are 2 groups now

6 shirts ==>they all have choices of 9 pants therefore total ways = \(6C1*9C1 = 54\)
NOW the left 2 shirts is going to avoid 3 pants therefore we have 6 pants hence total ways = \(2C1*6C1 =12\)

hence total choice = \(54 +12 =66\)

what you are doing is that you are just removing those shirts and pants.

hope it helps



I think you didn't get my question.

My question is that what is wrong in my method?

Rgds,
TGC !
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Re: Barbara has 8 shirts and 9 pants. How many clothing combinat  [#permalink]

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New post 07 Aug 2013, 00:15
targetgmatchotu wrote:
blueseas wrote:
targetgmatchotu wrote:
I have understood the above approaches . However, would like folks to answer that what is wrong with the below approach.

9 pants - 3 specific avoided - 6 pants
8 shirts - 2 specific avoided - 6 shirts

6C1 * 6C1 = 36 (select one from each for a combination)

What is wrong ?

Rgds,
TGC !


hi TGC,

Question is: Barbara has 8 shirts and 9 pants. How many clothing combinations does Barbara have, if she doesn’t wear 2 specific shirts with 3 specific pants?

it means he is going to wear all pants and shirts except few combinations he is going to avoid.
so there are 2 groups now

6 shirts ==>they all have choices of 9 pants therefore total ways = \(6C1*9C1 = 54\)
NOW the left 2 shirts is going to avoid 3 pants therefore we have 6 pants hence total ways = \(2C1*6C1 =12\)

hence total choice = \(54 +12 =66\)

what you are doing is that you are just removing those shirts and pants.

hope it helps



I think you didn't get my question.

My question is that what is wrong in my method?

Rgds,
TGC !


You are interpreting question in wrong way.

let say there are 9 pants numbered :1 2 3 4 5 6 7 8 9
AND 8 shirts numbered : 10 11 12 13 14 15 16 17

now question is saying you cannot wear pants numbered : 1, 2, 3 with shirt 10 and 11
but this doesnt mean that you cannot wear pants numbered : 1, 2, 3 with shirt 12 13 14 15 16 17 ===>you are missing this case.

what you are doing is that you are just removing pants 1,2,3 and shirts 10,11.
thats why you are having 6 shirts and 6 pants.

hope this now makes sense.
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GIVE VALUE TO OFFICIAL QUESTIONS...



GMAT RCs VOCABULARY LIST: http://gmatclub.com/forum/vocabulary-list-for-gmat-reading-comprehension-155228.html
learn AWA writing techniques while watching video : http://www.gmatprepnow.com/module/gmat-analytical-writing-assessment
: http://www.youtube.com/watch?v=APt9ITygGss

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Re: Barbara has 8 shirts and 9 pants. How many clothing combinat  [#permalink]

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Re: Barbara has 8 shirts and 9 pants. How many clothing combinat &nbs [#permalink] 09 Oct 2017, 21:25
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