# Ordering 29 CD's; how many ways?

• Mar 23rd 2012, 02:30 AM
timmeh
Ordering 29 CD's; how many ways?
In a library you find 12 Classical CD's, 10 Pop CD's and 7 Jazz CD's.

Question: In how many ways can you order those 29 CD's if the Classical CD's as well as the Pop CD's have to be next to each other?

How should I tackle this problem?

The answer should be: 6.3 * 10 ^ 20
• Mar 23rd 2012, 03:42 AM
Plato
Re: Ordering 29 CD's; how many ways?
Quote:

Originally Posted by timmeh
In a library you find 12 Classical CD's, 10 Pop CD's and 7 Jazz CD's.
Question: In how many ways can you order those 29 CD's if the Classical CD's as well as the Pop CD's have to be next to each other?
How should I tackle this problem? The answer should be: 6.3 * 10 ^ 20

I can see no reading of this question that would give that answer.
I get $(12!)(10!)(9!)~.$ Can you see how?
• Mar 23rd 2012, 04:37 PM
Soroban
Re: Ordering 29 CD's; how many ways?
Hello, timmeh!

Quote:

In a library there are 29 CD's: 12 Classical CD's, 10 Pop CD's and 7 Jazz CD's.

In how many ways can you order those 29 CD's if the Classical CD's as well as the Pop CD's have to be next to each other?

The answer should be: 6.3 * 10^20

It says: the 12 Classical CD's must be adjacent, and the 10 Pop CD's must be adjacent.

Duct-tape the 12 Classical CD's together.
Duct-tape the 10 Pop CD's together.

Then we have nine"CD's" to arrange:. $\boxed{\text{ Classical }}\;\boxed{\text{ Pop }}\;\boxed{J}\;\boxed{J}\; \boxed{J}\;\boxed{J}\; \boxed{J}\;\boxed{J}\;\boxed{J}$
These can be arranged in $9!$ ways.

The 12 Classical CD's can be arranged in $12!$ ways.
The 10 Pop CD's can be arranged in $10!$ ways.

Therefore, there are: . $9!\cdot12!\cdot10! \;\approx\; 6.3 \times 10^{20}$ ways.
• Mar 25th 2012, 05:34 AM
timmeh
Re: Ordering 29 CD's; how many ways?
Thanks Soroban, that makes it clear!