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Context: In January I will be taking a concurrent programming examination, part of which will involve calculating the number of interleavings based on a formula which divides two sets of factorials.

The problem I have is that calculators are prohibited, so it must be calculated on pen and paper, so I wonder what the best strategy is here?

The professor's suggestion was to simplify the factorial division, but the example he used didn't show how he simplified it, but also the numbers he was using were smaller.

One example I worked on is as follows:

12! / (6! x 6!)

The smallest I can simplify this to is as follows (although let me know if it can be simplified further as I may be wrong):

(2 * 11 * 10 * 3 * 4 * 7) \ (5 * 4)

This still seems like an excessive calculation for pen and paper (the question will be worth a single mark, also).

What would you suggest as a strategy for calculating the division of two factorials on pen and paper? (the goal being to come up with a strategy which takes the least amount of time to work out using pen and paper).

The result must be expressed as a single number, so using this example the result would be 924.


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up vote 4 down vote accepted

You can immediately cancel the $2\times10$ on the top with the $5\times4$ on the bottom, leaving $11\times3\times4\times7$.


The calculation shouldn't be too hard: $3\times4$ is easy, $12\times7$ is also pretty easy, and multiplication by $11$ is simple: If the two digits are $xy$, then $11\times xy=x(x+y)y$ (potentially carrying digits), which is easy to do mentally. For example, $11\times 23=253$. How big are these factorials going to be exactly?

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I think what the professor said was that the result would be no larger than 1,000 (or something to that effect), so there shouldn't be any excessive calculations. – Ciaran Gallagher Dec 19 '12 at 22:18
@CiaranG Then this'll probably be as complicated as it'll get. You have the right idea for simplifying it. Just note that since the result will always be an integer, you can always cancel enough to get rid of the denominator. Do that first and then just carry through the remaining multiplications. It might be a little tedious, but it shouldn't be too bad that way. – Robert Mastragostino Dec 20 '12 at 0:27

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