ro86 wrote:
1. Nine dogs are split into 3 groups to pull one of three sleds in a race. How many different assignments of dogs to sleds are possible?
2. In how many different ways can a group of 9 people be divided into 3 groups, with each group containing 3 people?
Are both the questions the same?
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 \(\frac{(mn)!}{(n!)^m*m!}\).
In the first case: I think the order is important, as we'll have group #1, #2 and #3 assigned to specific task. So we should use first formula, mn=9, m=3 groups n=3 objects (dogs):
\(\frac{(mn)!}{(n!)^m}=\frac{9!}{(3!)^3}=1680\).
This can be done in another way as well: \(9C3*6C3*3C3=1680\), (9C3 # of ways of choosing 3 from 9, 6C3 # of ways of choosing 3 from 6, 3C3 # of ways of choosing 3 from 3).
In the second case: I think the order is NOT important, as we won't have group #1, #2 and #3. So we should use second formula, again \(mn=9\), \(m=3\) groups \(n=3\) objects (people):
\(\frac{(mn)!}{(n!)^m*m!}=\frac{9!}{(3!)^3*3!}=280\).
This can be done in another way as well: \(\frac{9C3*6C3*3C3}{3!}=280\), we are dividing by \(3!\) as there are 3 groups and order doesn't matter.
how are you dividing groups in the first case with 3 members each?