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

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Intern
Joined: 25 Jan 2013
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Barbara has 8 shirts and 9 pants. How many clothing combinat  [#permalink]

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03 Mar 2013, 11:05
3
1
12
00:00

Difficulty:

25% (medium)

Question Stats:

70% (01:14) correct 30% (01:34) wrong based on 158 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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03 Mar 2013, 12:08
3
1
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
##### General Discussion
Intern
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Re: Barbara has 8 shirts and 9 pants. How many clothing combinat  [#permalink]

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Updated on: 04 Apr 2013, 00:03
1
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, 03:57.
Last edited by michellark on 04 Apr 2013, 00:03, 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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12 Mar 2013, 04:21
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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06 Aug 2013, 21:30
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 !
Director
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Re: Barbara has 8 shirts and 9 pants. How many clothing combinat  [#permalink]

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06 Aug 2013, 21:44
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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Director
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Re: Barbara has 8 shirts and 9 pants. How many clothing combinat  [#permalink]

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07 Aug 2013, 01:03
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 !
Director
Joined: 14 Dec 2012
Posts: 688
Location: India
Concentration: General Management, Operations
GMAT 1: 700 Q50 V34
GPA: 3.6
Re: Barbara has 8 shirts and 9 pants. How many clothing combinat  [#permalink]

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07 Aug 2013, 01:15
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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Re: Barbara has 8 shirts and 9 pants. How many clothing combinat  [#permalink]

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24 Nov 2019, 18:57
antoxavier wrote:
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

If we exclude the 2 specific shirts and 3 specific pants that she won’t wear together as an outfit, there are 6 x 6 = 36 possible outfits.

If we exclude the 2 specific shirts, then any of the remaining 6 shirts can pair with the 3 specific pants. So there are 6 x 3 = 18 such outfits.

If we exclude the 3 specific pants, then any of the remaining 6 pants can pair with the 2 specific shirts. So there are 6 x 2 = 12 such outfits.

Therefore, there are a total of 36 + 18 + 12 = 66 possible outfits.

Alternate Solution:

Without any restrictions, she has 8 x 9 = 72 possible outfits. We are told that Barbara does not wear 2 specific shirts with 3 specific pants; thus, of the 72 total outfits, 2 x 3 = 6 of them are never worn. So, the number of possible outfits is 72 - 6 = 66.

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Intern
Joined: 27 Nov 2019
Posts: 5
Re: Barbara has 8 shirts and 9 pants. How many clothing combinat  [#permalink]

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27 Nov 2019, 06:43
The right answer is B.66 clothing combinat
Re: Barbara has 8 shirts and 9 pants. How many clothing combinat   [#permalink] 27 Nov 2019, 06:43
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