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Barbara has 8 shirts and 9 pants. How many clothing combinat
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03 Mar 2013, 11:05
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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
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03 Mar 2013, 12:08
Total no of combinations available is= 8 shirts X 9 pants = 72 Eliminate the 2shirts X 3 pants combo = 6 which gives.......>> 726=66




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Re: Barbara has 8 shirts and 9 pants. How many clothing combinat
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Updated on: 04 Apr 2013, 00:03
Clothing combination of shirt and pant left are 36 i.e shirt=82=6 pant=93=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
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12 Mar 2013, 04:21
8 shirts 9 pants
we can combine 2 shirts with (93) 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
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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 !



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Re: Barbara has 8 shirts and 9 pants. How many clothing combinat
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06 Aug 2013, 21:44
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 now6 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
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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 now6 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
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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 now6 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
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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. Answer: B
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Re: Barbara has 8 shirts and 9 pants. How many clothing combinat
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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
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27 Nov 2019, 06:43






