Bunuel wrote:
GENERAL RULE:
1. The number of ways in which mn different items can be divided equally into m groups, each containing n objects and the order of the groups is important is \(\frac{(mn)!}{(n!)^m}\)
2. The number of ways in which mn different items can be divided equally into m groups, each containing n objects and the order of the groups is NOT important is \(\frac{(mn)!}{(n!)^m * m!}\)
I don't think that's right. For example, let's find out in how many ways 4 different items can be split into 2 groups of 2, when the order is NOT important, using the 2nd formula. \(\frac{(mn)!}{(n!)^m*m!}\) = \(\frac{(4)!}{(2!)^2*2!}\) = 3
This result can be disproved very easily graphically:
1. Draw a square
2. Draw an x through the square drawing 2 lines, corner to corner
3. Represent the 4 objects on each corner of the square
4. Number the lines that link all the square's corners together
There are 6 lines which represent all 6 possible groups of 2 objects; every object is connected by a line to all other objects a single time, forming 6 possible groups of 2.
The answer is 6. Another solution is 4C2 * 2C2 = 6
nCr = \(\frac{(n!)}{(r!(n-r)!)}\) = \(\frac{(4!)}{(r!(4-2)!)}\) * \(\frac{(2!)}{(r!(2-2)!)}\) = 6
The groups have not been ordered yet and there is no need to divide by anything else; we proved graphically that 6 is the answer so dividing by 2! or anything else would lead to an incorrect result.
If the order was important, there would be \(6!/(6-2)! = 30\) ways to organize 4 objects into 2 groups of 2.[/b]
There would be 2 positions to fill with 6 possible groups. 6 possible groups in the 1st position and 5 possible groups in the 2nd position.
nPr= \(\frac{n!}{(n-r)!}\) = \(\frac{6!}{(6-2)!}\) = 6 * 5 = 30
The 1st formula actually represents the number of ways in which the order is NOT important\(\frac{(mn)!}{(n!)^m}\) = \(\frac{(4)!}{(2!)^2}\) = 6
THE ORIGINAL QUESTION:The order is not important in this question.nCr = \(\frac{(n!)}{(r!(n-r)!)}\) = \(\frac{(9!)}{(3!(9-3)!)}\) * \(\frac{(6!)}{(3!(6-3)!)}\) * \(\frac{(3!)}{(3!(3-3)!)}\) = 1680
or, alternatively:
\(\frac{(mn)!}{(n!)^m}\) = \(\frac{(9)!}{(3!)^3}\) = 1680
If the order mattered:nPr= \(\frac{n!}{(n-r)!}\) = \(\frac{1680!}{(1680-3)!}\) = 1680 * 1679 * 1678 = 4733168160