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Algebra
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28 May 2007 18:36:39 IST
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Nano0101 u messed up. Observe the following steps:
3 > 2 ........... eqn. 1
1 < 5..............eqn. 2
Now u just cannot multiply eqns 1 & 2 to say that 3 > 10. It is not always true as u can see in the above e.g. The bottom line is that,
a > b
c < d
No one in the world can predict that ac > bd.
So ur step involving the following is not valid.
(300!/100!) >100200 -------1
100! < 100100 ----------2
300! > 100300 --------from 1 n 2
100! < 100100 ----------2
300! > 100300 --------from 1 n 2
But nice try yaar !!!
Now Prathima, I'm not convinced by ur method of approach but it was a good attempt altogether. Can u think of anything else ?
30 May 2007 21:30:10 IST
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Use this approach
Stirling's Approximation for n!
When evaluating distribution functions for statistics, it is often necessary to evaluate the factorials of sizable numbers, as in the binomial distribution:

A helpful and commonly used approximate relationship for the evaluation of the factorials of large numbers is Stirling's approximation:

A slightly more accurate approximation is the following

but in most cases the difference is small. This additional term does give a way to assess whether the approximation has a large error.
Stirling's approximation is also useful for approximating the log of a factorial, which finds application in evaluation of entropy in terms of multiplicity, as in the Einstein solid. The log of n! is

but the last term may usually be neglected so that a working approximation is















in both the numbers it is the product of 300 numbers
in 300! 200 numbers r greater than the 200 no. in 100^300
only 100 no. r less
so i think 300! is greater than 100^300
do correct me if i m wrong