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.
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