document.write( "Question 1081665: Simplify (n+5)!/(n+3)! \n" ); document.write( "
Algebra.Com's Answer #695716 by jim_thompson5910(35256)\"\" \"About 
You can put this solution on YOUR website!
By definition, \r
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\n" ); document.write( "\n" ); document.write( "n! = n*(n-1)*(n-2)*...*3*2*1\r
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\n" ); document.write( "\n" ); document.write( "Whatever value you pick for n, you count your way down til you get to 1. All of the integers in the countdown are multiplied.\r
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\n" ); document.write( "\n" ); document.write( "Using similar logic, the expression (n+5)! can be expanded as such\r
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\n" ); document.write( "\n" ); document.write( "(n+5)! = (n+5)*(n+4)*(n+3)*(n+2)*(n+1)*...*3*2*1\r
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\n" ); document.write( "\n" ); document.write( "and that can be turned into\r
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\n" ); document.write( "\n" ); document.write( "(n+5)! = (n+5)*(n+4)*(n+3)!\r
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\n" ); document.write( "\n" ); document.write( "notice how the red portions match up. They are equal expressions. In other words,\r
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\n" ); document.write( "\n" ); document.write( "(n+3)! = (n+3)*(n+2)*(n+1)*...*3*2*1\r
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\n" ); document.write( "\n" ); document.write( "Let's use that idea to simplify\r
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\n" ); document.write( "\n" ); document.write( "\"%28%28n%2B5%29%21%29%2F%28%28n%2B3%29%21%29=%28n%2B5%29%28n%2B4%29\"\r
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\n" ); document.write( "\n" ); document.write( "\"%28%28n%2B5%29%21%29%2F%28%28n%2B3%29%21%29=n%5E2%2B9n%2B20\"\r
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\n" ); document.write( "\n" ); document.write( "Final Answer: \"n%5E2%2B9n%2B20\"
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