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Ten different books, including 2 Portuguese books and 3 Math

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Ten different books, including 2 Portuguese books and 3 Math [#permalink] New post 24 Apr 2007, 13:11
Ten different books, including 2 Portuguese books and 3 Math books, must be placed on a stand, in any order. However, the 2 Portuguese books must be together, and the 3 Math books must be together. The number of different ways to organize the books is:

(A) 3628800
(B) 60480
(C) 5040
(D) 2520
(E) 1440
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 [#permalink] New post 24 Apr 2007, 13:59
There are 10 books.
2 Portuguese books and 3 Maths books should be placed together. We can consider them as unit. So one is Portuguese Unit and another one is Maths unit. Every books other than Portuguese and Maths will make a unit.
In total we have to arrange the 7 units.
Number of ways to arrange 7 units = 7! = 5040.
Number of ways to arrange the Portuguese books = 2!
Number of ways to arrange the Maths books = 3!

So total number of ways to arrange in order required = 7! * 3! * 2! = 60480.

Whats the OA

Last edited by vshaunak@gmail.com on 24 Apr 2007, 15:00, edited 1 time in total.
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 [#permalink] New post 24 Apr 2007, 14:51
vshaunak@gmail.com wrote:
There are 10 books.
2 Portuguese books and 3 Maths books should be placed together. We can consider them as unit. So one is Portuguese Unit and another one is Maths unit. Every books other than Portuguese and Maths will make a unit.
In total we have to arrange the 7 units.
Number of ways to arrange 7 units = 7! = 5040.


Ditto for me
but the 3 Maths books can be arranged amongst themselves and so can the Portugese ones.
thats 3! ways for the Math books and
2! ways for the Portugese books

Total = 7!3!2! = 60480..........(B) for me
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 [#permalink] New post 24 Apr 2007, 14:57
oops.....I forgot to take that into account. Thanks Vikramjit...
I edit my post.
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 [#permalink] New post 24 Apr 2007, 23:38
Are we considering the math/portugese books from being different from one another within the classification ?

If we dont, then answer = 7! = 5040

If we do, then 7! x 3! x 2! = 60480

So what should we do here ?

Gen I have seen problems where say letters of an alphabet needs to be re-arranged with certain constraints like A,D,F should always be together.
Here we multiply by 3! taking into consideration the permutation among A,D, and F (they are distinct alphabets)

Any inputs ?
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 [#permalink] New post 26 Apr 2007, 08:45
grad_mba wrote:
Are we considering the math/portugese books from being different from one another within the classification ?

If we dont, then answer = 7! = 5040

If we do, then 7! x 3! x 2! = 60480

So what should we do here ?

Gen I have seen problems where say letters of an alphabet needs to be re-arranged with certain constraints like A,D,F should always be together.
Here we multiply by 3! taking into consideration the permutation among A,D, and F (they are distinct alphabets)

Any inputs ?

It's like this
If you were to arrange A, B, C, D in as many different ways as possible then your answer would be 4!
but if you were to arrange them in such a way that A & B were always together then you would have
AB C D
that's 3!*2!
3! cos you have 3 units now ->AB, C & D
2! cos A & B can be arranged within themselves.
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 [#permalink] New post 26 Apr 2007, 11:01
Thanks Vikram, but guess my question was different..

we multiply by 2! becasue A is different from B

in the given problem, are we supposed to consider a given math book different from the other math books ?
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 [#permalink] New post 26 Apr 2007, 11:41
5 books that can be arranged without regard to order:

10C5 = 30,240

Math and Portuguese arranged with regard to order:

10P5 = 30,240

Total = 30,240 + 30,240 = 60,480

B
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 [#permalink] New post 27 Apr 2007, 13:41
Quote:
Total = 7!3!2! = 60480..........(B) for me


that is enough..
  [#permalink] 27 Apr 2007, 13:41
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Ten different books, including 2 Portuguese books and 3 Math

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