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hey,
Looks like the question is too easy for further explanation. However, i could not understand how we got that answer.
Pleaseeee reply.
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Bunuel
bibha
hey,
Looks like the question is too easy for further explanation. However, i could not understand how we got that answer.
Pleaseeee reply.

In how many different ways can 4 physics, 2 math and 3 chemistry books be arranged in a row so that all books of the same branch are together?

A. 1242
B. 1728
C. 1484
D. 1734
E. 1726

There are three branches, three units of books: {physics}{math}{chemistry} - aranging branches 3!;

Arranging the books within the branches:
physics - 4!;
math - 2!;
chemistry - 3!;

Total: 3!*4!*2!*3!.

Answer: B.


Hi Bunuel,
I apologize for opening a very old post but had to ..
my doubt is if we are arranging 4 physics books then doesn't the arrangement will be 4!/4!, similarly math and chemistry..and over all arrangement should be just 3!..please explain where I am missing..
Thank you,
Vivek.
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viveksingh222
Bunuel
bibha
hey,
Looks like the question is too easy for further explanation. However, i could not understand how we got that answer.
Pleaseeee reply.

In how many different ways can 4 physics, 2 math and 3 chemistry books be arranged in a row so that all books of the same branch are together?

A. 1242
B. 1728
C. 1484
D. 1734
E. 1726

There are three branches, three units of books: {physics}{math}{chemistry} - aranging branches 3!;

Arranging the books within the branches:
physics - 4!;
math - 2!;
chemistry - 3!;

Total: 3!*4!*2!*3!.

Answer: B.


Hi Bunuel,
I apologize for opening a very old post but had to ..
my doubt is if we are arranging 4 physics books then doesn't the arrangement will be 4!/4!, similarly math and chemistry..and over all arrangement should be just 3!..please explain where I am missing..
Thank you,
Vivek.

It would be so if we were told that books in each branch are identical. But that's not our case, for example, out of 2 math books one could be on algebra and another on arithmetic thus they can be arranged within math branch as {algebra}{arithmetic} or {arithmetic }{algebra}.

Hope it's clear.
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Hi Bunuel !!

I am not sure why you have assumed that the books within one subject are actually different !
I believe the books can also be same , solution in that case should be 3!.

You have assumed the same in this problem :

In how many ways can 11 books on English and 9 books on French be placed in a row on a shelf so that two books on French may not be together?

in-how-many-ways-can-11-books-on-english-and-9-books-on-87352.html#p1394032

Your solution was :
I would offer different solution.

We have 11 English and 9 French books, no French books should be adjacent.

Imagine 11 English books in a row and empty slots like below:

*E*E*E*E*E*E*E*E*E*E*E*

Now if 9 French books would be placed in 12 empty slots, all French books will be separated by English books.

So we can "choose" 9 empty slots from 12 available for French books, which is 12C9=220.



I believe in both these questions we should consider the books to be different. But I am confused because of the two opposite approaches applied to 2 similar problems. Where should one draw the line.

Please have a look . Thanks in advance !!!!
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Ravshonbek
In how many different ways can 4 physics, 2 math and 3 chemistry books be arranged in a row so that all books of the same branch are together?

A. 1242
B. 1728
C. 1484
D. 1734
E. 1726


4! x 2! x 3! x 3! = 1728

The last 3! came for the number of ways Physics, Math and Chemistry are arranged.
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solitaryreaper
Hi Bunuel !!

I am not sure why you have assumed that the books within one subject are actually different !
I believe the books can also be same , solution in that case should be 3!.

You have assumed the same in this problem :

In how many ways can 11 books on English and 9 books on French be placed in a row on a shelf so that two books on French may not be together?

in-how-many-ways-can-11-books-on-english-and-9-books-on-87352.html#p1394032

Your solution was :
I would offer different solution.

We have 11 English and 9 French books, no French books should be adjacent.

Imagine 11 English books in a row and empty slots like below:

*E*E*E*E*E*E*E*E*E*E*E*

Now if 9 French books would be placed in 12 empty slots, all French books will be separated by English books.

So we can "choose" 9 empty slots from 12 available for French books, which is 12C9=220.



I believe in both these questions we should consider the books to be different. But I am confused because of the two opposite approaches applied to 2 similar problems. Where should one draw the line.

Please have a look . Thanks in advance !!!!

Thanks!
Exact same query? When do we have to consider the books different and when same , if it is not explicitly stated?
Experts please answer... TIA!
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tushain
solitaryreaper
Hi Bunuel !!

I am not sure why you have assumed that the books within one subject are actually different !
I believe the books can also be same , solution in that case should be 3!.

You have assumed the same in this problem :

In how many ways can 11 books on English and 9 books on French be placed in a row on a shelf so that two books on French may not be together?

in-how-many-ways-can-11-books-on-english-and-9-books-on-87352.html#p1394032

Your solution was :
I would offer different solution.

We have 11 English and 9 French books, no French books should be adjacent.

Imagine 11 English books in a row and empty slots like below:

*E*E*E*E*E*E*E*E*E*E*E*

Now if 9 French books would be placed in 12 empty slots, all French books will be separated by English books.

So we can "choose" 9 empty slots from 12 available for French books, which is 12C9=220.



I believe in both these questions we should consider the books to be different. But I am confused because of the two opposite approaches applied to 2 similar problems. Where should one draw the line.

Please have a look . Thanks in advance !!!!

Thanks!
Exact same query? When do we have to consider the books different and when same , if it is not explicitly stated?
Experts please answer... TIA!

Because then it would be unsolvable. GMAT questions ideally state this explicitly.
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Ravshonbek
In how many different ways can 4 physics, 2 math and 3 chemistry books be arranged in a row so that all books of the same branch are together?

A. 1242
B. 1728
C. 1484
D. 1734
E. 1726

Total books: 4 Math, 3 English, 2 Analytical Ability
Assume the books of each subject to be a bundle
Hence we have 3 bundles of Math, English and Analytical Ability

Total ways to arranging 3 bundles = 3!
Books inside each bundle can also be arranged.

Total ways = 3!*4!*3!*2! = 6*24*6*2 = 36*48 = 1728
By paying attention to the options, we can reach the correct one without calculating
As the answer would be a number rending with the digit 8 and there is only one such number in the given options.

Correct Option: B
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Ravshonbek
In how many different ways can 4 physics, 2 math and 3 chemistry books be arranged in a row so that all books of the same branch are together?

A. 1242
B. 1728
C. 1484
D. 1734
E. 1726


4 physics book can be arranged among themselves in 4! ways
Similarly, 3 chemistry and 2 math in 3! and 2! ways.

Now considering them as 3 groups, the 3 groups can be arranged in 3! ways.

so the total number of ways = 4! * 3! * 2! * 3!
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I am pretty sure this is a stupid question but here it goes:

Each category of books can be arranged the following number of ways.
physics 4!; math 2!; chemistry 3!;

there are three categories, so the number of ways which different categories can be arranged is 3!.
Now, here's my problem. Bunuel

I thought it would be (3!+4!+2!)*3!, because these are number of ways which the books of different categories are being arranged.

Why should they be multiplied?

Thank you!!
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HappyQuakka
I am pretty sure this is a stupid question but here it goes:

Each category of books can be arranged the following number of ways.
physics 4!; math 2!; chemistry 3!;

there are three categories, so the number of ways which different categories can be arranged is 3!.
Now, here's my problem. Bunuel

I thought it would be (3!+4!+2!)*3!, because these are number of ways which the books of different categories are being arranged.

Why should they be multiplied?

Thank you!!

Say the order is {physics: 1, 2, 3, 4}{math: 1, 2}{chemistry: 1, 2, 3}

For two ordering of say {math: 1, 2}: {1, 2} and {2, 1} we'll have:
{physics: 1, 2, 3, 4}{math: 1, 2}{chemistry: 1, 2, 3}
{physics: 1, 2, 3, 4}{math: 2, 1}{chemistry: 1, 2, 3}

Similarly for all 3! orderings of {chemistry: 1, 2, 3} we'll have 6 times as many for each above 2!*3! and for 4! orderings of {physics: 1, 2, 3, 4} we'll have 24 as many as previous number, so 2!*3!*4!.

More generally, we multiply because of FUNDAMENTAL PRINCIPLE OF COUNTING, which states if an operation can be performed in ‘m’ ways and when it has been performed in any of these ways, a second operation that can be performed in ‘n’ ways then these two operations can be performed one after the other in ‘m*n’ ways.
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Ravshonbek
In how many different ways can 4 physics, 2 math and 3 chemistry books be arranged in a row so that all books of the same branch are together?

A. 1242
B. 1728
C. 1484
D. 1734
E. 1726

We can illustrate the arrangement as follows:

[P-P-P-P] [M-M] [C-C-C]

So, the 3 groups of books can be arranged in 3! = 6 ways. Furthermore, the physics books can be arranged in 4! = 24 ways, the math books in 2! = 2 ways, and the chemistry books in 3! = 6 ways.

Thus, the total number of ways to arrange the books is 6 x 24 x 2 x 6 = 1728 ways.

Answer: B
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Bunuel

bibha
hey,
Looks like the question is too easy for further explanation. However, i could not understand how we got that answer.
Pleaseeee reply.
In how many different ways can 4 physics, 2 math and 3 chemistry books be arranged in a row so that all books of the same branch are together?

A. 1242
B. 1728
C. 1484
D. 1734
E. 1726

There are three branches, three units of books: {physics}{math}{chemistry} - aranging branches 3!;

Arranging the books within the branches:
physics - 4!;
math - 2!;
chemistry - 3!;

Total: 3!*4!*2!*3!.

Answer: B.
­Hello! Wonder how to calculate 3!*4!*2!*3! this quickly?
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lnyngayan

Bunuel

bibha
hey,
Looks like the question is too easy for further explanation. However, i could not understand how we got that answer.
Pleaseeee reply.
In how many different ways can 4 physics, 2 math and 3 chemistry books be arranged in a row so that all books of the same branch are together?

A. 1242
B. 1728
C. 1484
D. 1734
E. 1726

There are three branches, three units of books: {physics}{math}{chemistry} - aranging branches 3!;

Arranging the books within the branches:
physics - 4!;
math - 2!;
chemistry - 3!;

Total: 3!*4!*2!*3!.

Answer: B.
­Hello! Wonder how to calculate 3!*4!*2!*3! this quickly?

­3!*4!*2!*3! = 6*24*2*6 = something with the units digit of 8.

Or just calculate 6*24*2*6 = 72*24.
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