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For every integer n greater than 1, n! is defined to be the product of

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For every integer n greater than 1, n! is defined to be the product of [#permalink] New post   27 Sep 2020, 00:58
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For every integer n greater than 1, n! is defined to be the product of the first n positive integers. For example, 4!=1x2x3x4=24. which of the following is equal to (8!)(7!)?

A) 56
B) 8x7!
C) 7x8!
D) 8x(7!)^2
E) (56!)^2
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D
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Re: For every integer n greater than 1, n! is defined to be the product of [#permalink] New post   27 Sep 2020, 07:37
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n! is n * (n - 1) * (n - 2) * .... * 1. Therefore n! can be written as n * (N - 1)! or n*(n - 1) * (n - 2)! and so on.

8! is 8 * 7 * 6 *...*1 = 8 * 7!

8! * 7! = 8 * 7! * 7! = \(8 * 7!^2\)


Option D

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Re: For every integer n greater than 1, n! is defined to be the product of [#permalink] New post   27 Sep 2020, 08:13
[quote="hkhanexam"]For every integer n greater than 1, n! is defined to be the product of the first n positive integers. For example, 4!=1x2x3x4=24. which of the following is equal to (8!)(7!)?

(8!)(7!) = 8*(7!)(7!) = 8*(7!)^2

IMO D
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Re: For every integer n greater than 1, n! is defined to be the product of [#permalink] New post   27 Sep 2020, 08:21
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hkhanexam wrote:
For every integer n greater than 1, n! is defined to be the product of the first n positive integers. For example, 4!=1x2x3x4=24. which of the following is equal to (8!)(7!)?

A) 56
B) 8x7!
C) 7x8!
D) 8x(7!)^2
E) (56!)^2


8! is 8*7! and 7!
so we get 8*(7!)^2
option D
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Re: For every integer n greater than 1, n! is defined to be the product of [#permalink] New post   02 Sep 2022, 22:25
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Given that n! is defined to be the product of the first n positive integers and we need to find which of the following is equal to (8!)(7!)

8! * 7! = 1*2*3..*6*7*8 * 7! = (1*2*3..*6*7)*8 * 7! = 7! * 8 * 7! = 8 * (\(7!^2\))

So, Answer will be D.
Hope it helps!
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Re: For every integer n greater than 1, n! is defined to be the product of [#permalink]
02 Sep 2022, 22:25
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