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Ten different books, including 2 Portuguese books and 3 Math [#permalink]
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:

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.

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

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)

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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