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Simplify the expression \(√7!\), where \(n!\) stands for \(n*(n-1)*(n-2).... 2*1\).

 Apr 18, 2023
edited by Keihaku  Apr 18, 2023
 #1
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7! = 5040 = 2^4 * 3^2 * 5 * 7

 

Sqrt{2^4 * 3^2 * 5 * 7]= 2^2 * 3 * sqrt(5 *7) = 12 sqrt(35) - which is what you want.

 Apr 18, 2023
 #2
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Thank you, guest, but i'm sorry that's incorrect! 

Keihaku  Apr 18, 2023
 #3
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sqrt(7!) simplifies to 4*sqrt(210).

 Apr 18, 2023
 #4
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Sorry, thats incorrect too! I'm not sure if its the same guest, but thanks for trying again!

Keihaku  Apr 18, 2023
 #5
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To calculate the square root of 7! where n! stands for n*(n-1)(n-2)...2*1, we can use the factorial property:

7! = 7654321

sqrt(7!) = sqrt(7654321) = sqrt(7) * sqrt(6) * sqrt(5) * sqrt(4) * sqrt(3) * sqrt(2) * sqrt(1)

sqrt(7) * sqrt(6) * sqrt(5) * sqrt(4) * sqrt(3) * sqrt(2) * sqrt(1) = 7^(1/2) * 6^(1/2) * 5^(1/2) * 4^(1/2) * 3^(1/2) * 2^(1/2) * 1

 

Using this information to jumpstart your question, try simplifying the last expression.

 Apr 18, 2023

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