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![[Post New]](/templates/default/images/icon_minipost_new.gif) 28 May 2007 15:44:52 IST
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Which is greater among 100 300 and 300 ! ?? Please give the solution & not just the answer. I am waiting for a magnificient solution.
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You never know what is enough till you know what is more than enough.
Titun |
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 28 May 2007 16:13:03 IST
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i think 300! is greater 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
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GBXDHNXDFNHBHRSDRGWEASGSEDH |
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 28 May 2007 17:35:13 IST
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100 300 300 ! = 100! (300!/100!) (300!/100!) has 200 terms which are greater than 100 => (300!/100!) >100200 -------1 100! < 100100 ----------2 300! > 100300 --------from 1 n 2
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In the process of learnin..............blunders do happen !!! |
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 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 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 ?
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You never know what is enough till you know what is more than enough.
Titun |
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 29 May 2007 10:11:40 IST
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300! > 100300
This is a direct consequence of Stirling's approximation of factorial values.
Stirling's approximation for higher factorials states that
 There is also another relation
(n/3)n < n! < (n/2)n
That solves the Problem!
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Destiny is what you make
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 29 May 2007 21:11:19 IST
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good work amaron
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 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
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The Scientist does not study nature because it is useful; he studies it because he delights in it, & he delights in it because it is beautiful. If nature were not beautiful, it would not be worth knowing, life would not be worth living. Ofcourse I do not here speak of that beauty that strikes the senses, the beauty of qualities & appearances; not that I undervalue such beauty, far from it, but it has nothing to do with science; I mean that profounder beauty which comes from the harmoniuos order of the parts, & which a pure intelligence can grasp. |
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