No of Schedules

Question

Given,                       

Transaction T1 has n operation                                                            

Transaction T2 has m operation

Prove that, the total number of schedules possible is  $\frac{(m+n)!}{m!n!}$

Comments

its like distribute m+n different objects to 2 unlabeled boxes ..

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Also an example of unordered partition.

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yes both are same thing

Answer 1

First of all, let's permute everything; All operations of both the transactions.
Doing this will result in - (m+n)! 

Now,
We can't change the internal order of execution of operations within a schedule, we can just interleave the operations of both the transactions.

Therefore, to remove the permutations of internal operations of transaction T1, we need to divide the above expression by n!. Same goes for the transaction T2. Thus we get the expression $\frac{(m+n)!}{m!n!}$