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Re: A restaurant is hanging 7 large tiles on its wall in a single row [#permalink]
vikasp99 wrote:
A restaurant is hanging 7 large tiles on its wall in a single row. How many arrangements of tiles are possible if there are 3 white tiles and 4 blue tiles?

A. 35
B. 12
C. 21
D. 84
E. 28



Hello Bunuel,

Isn't this question confusing. 3 White tiles could be different white tiles and/or 4 blue tiles could be different blue tiles ? How we can make sure that all white/blue tiles are similar ?

Thanks,
Amm
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Re: A restaurant is hanging 7 large tiles on its wall in a single row [#permalink]
ammuseeru
I understand your attention to detail (based on some convoluted and tricky gmat questions), but it I have not found it necessary for combination questions. Most people (including and especially me) find this section confusing. The questions ask straight forward questions and simply require application of rules. This question tests 'permutation with repeated elements'.

\(P(n; n_1, n_2, . . . , n_k) = \frac{n!}{n_1! n_2! . . . n_k!}\)
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Re: A restaurant is hanging 7 large tiles on its wall in a single row [#permalink]
chetan2u wrote:
vikasp99 wrote:
A restaurant is hanging 7 large tiles on its wall in a single row. How many arrangements of tiles are possible if there are 3 white tiles and 4 blue tiles?

A. 35
B. 12
C. 21
D. 84
E. 28


Hi..

Total ways these 7 tiles can be arranged =7!
However these contain duplicity as 3 tiles are of one colour and other 4 are of other colour..
So eliminating these = \(\frac{7!}{3!4!}\)=35

 

­Hi there, isn't it a combination problem, as the question stem doesn't express any nessacity of following certain order?
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Re: A restaurant is hanging 7 large tiles on its wall in a single row [#permalink]
Expert Reply
ShuvoDhk wrote:
chetan2u wrote:
vikasp99 wrote:
A restaurant is hanging 7 large tiles on its wall in a single row. How many arrangements of tiles are possible if there are 3 white tiles and 4 blue tiles?

A. 35
B. 12
C. 21
D. 84
E. 28

Hi..

Total ways these 7 tiles can be arranged =7!
However these contain duplicity as 3 tiles are of one colour and other 4 are of other colour..
So eliminating these = \(\frac{7!}{3!4!}\)=35



 

­Hi there, isn't it a combination problem, as the question stem doesn't express any nessacity of following certain order?

­Can you please clarify what you mean by "it's a combination problem"? What, according to you, should the answer be?

P.S. The wording would have been better if the question mentioned that 3 white tiles are identical and 4 blue tiles are also identical. If they are not, then the answer would simply be 7!.­
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Re: A restaurant is hanging 7 large tiles on its wall in a single row [#permalink]
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