I have a question about a corollary from a proof from Higher Algebra by Hall and knight.
146. To find the number of ways in which m+n things can be divided into groups of m and n things respectively.
Proof: This is equivalent to finding the number of combinations of m+n things m at a time, for every time we select one group of m things we leave a group of n things behind. Thus the required number is
(mm+n)=m!n!(m+n)!
Corollary. If n=m, the groups are equal, and in this case, the number of different ways of subdivision is
m!m!2!(2m)!;
for in any one way, it is possible to interchange the two groups without obtaining a new distribution.
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To understand the corollary provided above, I've tried to delve into the reasoning step-by-step.
Let's consider when n=m. The total number of things is 2m. We divide these 2m things into two groups, each containing m things. Applying the same logic as above this is equivalent to choosing m things out of m+m things because the remaining m things will automatically form the second group. We use the binomial coefficient to find the number of ways to choose m things out of 2m things.
(m2m)=m!m!(2m)!;
Suppose we denote the equal objects as m1 and m2 such that m1=m2=m.
The division by 2 accounts for the fact that
m1!m1!(2m1)!=m2!m2!(2m2)!
Since m1 and m2 are equal in size, swapping them does not result in a new unique division. This means each unique division of the 2m things into two equal groups is counted twice in the binomial coefficient calculation. Therefore, the number of unique ways to divide 2m things into two groups of m things each is:
m!m!2!(2m)!;
I apologize if this seems trivial, but I would really appreciate it if you could confirm whether my reasoning is correct.
146. To find the number of ways in which m+n things can be divided into groups of m and n things respectively.
Proof: This is equivalent to finding the number of combinations of m+n things m at a time, for every time we select one group of m things we leave a group of n things behind. Thus the required number is
(mm+n)=m!n!(m+n)!
Corollary. If n=m, the groups are equal, and in this case, the number of different ways of subdivision is
m!m!2!(2m)!;
for in any one way, it is possible to interchange the two groups without obtaining a new distribution.
...............................................................................................................................................................................................................................................................................................................................
To understand the corollary provided above, I've tried to delve into the reasoning step-by-step.
Let's consider when n=m. The total number of things is 2m. We divide these 2m things into two groups, each containing m things. Applying the same logic as above this is equivalent to choosing m things out of m+m things because the remaining m things will automatically form the second group. We use the binomial coefficient to find the number of ways to choose m things out of 2m things.
(m2m)=m!m!(2m)!;
Suppose we denote the equal objects as m1 and m2 such that m1=m2=m.
The division by 2 accounts for the fact that
m1!m1!(2m1)!=m2!m2!(2m2)!
Since m1 and m2 are equal in size, swapping them does not result in a new unique division. This means each unique division of the 2m things into two equal groups is counted twice in the binomial coefficient calculation. Therefore, the number of unique ways to divide 2m things into two groups of m things each is:
m!m!2!(2m)!;
I apologize if this seems trivial, but I would really appreciate it if you could confirm whether my reasoning is correct.
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