document.write( "Question 858584: How can 5000! (a large factorial) be found numerically?
\n" ); document.write( "Someone, call him Igor for this problem :) Anyway - wants to know how
\n" ); document.write( "many groups of 50 are there in 5000.
\n" ); document.write( "I started working this using nCr = n! / r!(n-r)!
\n" ); document.write( "nCr = 5000C50, and chose to settle for a ball-park (estimated figure) quote,
\n" ); document.write( "yet have no idea although the number C must lie between astronomical and ∞.
\n" ); document.write( "50! is solvable, but was wondering if 5000! can be partitioned or reduced
\n" ); document.write( "so a hand calculator could work it. Any shortcuts here would be appreciated!
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Algebra.Com's Answer #517285 by KMST(5328)\"\" \"About 
You can put this solution on YOUR website!
You do not need to calculate a number as astronomical as 5000!,
\n" ); document.write( "because ,
\n" ); document.write( "and that is slightly less astronomical.
\n" ); document.write( "According to the combinations function in the Excel spreadsheet program in my computer, 5000C50 is \"2.2839%2A10%5E120\" .
\n" ); document.write( "That must also an estimate, because there is no way it could calculate all 120 decimal places.
\n" ); document.write( "I also calculated 5000C50 as
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\n" ); document.write( "Excel also calculated \"5000%2A4999%2A4998%2A%22...%22%2A4953%2A4952%2A4951=6.9432%2A10%5E184\" for me.
\n" ); document.write( "I do not know if you can do that with a calculator.
\n" ); document.write( "A crude approximation of \"5000%21%2F4950%21=5000%2A4999%2A4998%2A%22...%22%2A4953%2A4952%2A4951\"
\n" ); document.write( "would be .
\n" ); document.write( "A better approximation for
\n" ); document.write( "
\n" ); document.write( "would be \"5000%2A4975%5E49=6.9475%2A10%5E184\" .
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