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If x, y, and z are positive integers, is (x!)(y!) > z! ?

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Bunuel
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If x, y, and z are positive integers, is (x!)(y!) > z! ? [#permalink] New post  05 Apr 2016, 08:31
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If x, y, and z are positive integers, is (x!)(y!) > z! ?

(1) x^y > z
(2) x + y = z
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chetan2u
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Re: If x, y, and z are positive integers, is (x!)(y!) > z! ? [#permalink] New post  05 Apr 2016, 10:09
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Bunuel wrote:
If x, y, and z are positive integers, is (x!)(y!) > z! ?

(1) x^y > z
(2) x + y = z



Hi,

lets see the choices and substitue values to see answer

(1) \(x^y > z\)
x = 3, y=2 and z=6... 3!2!>6!-- NO
x=3, y=2 and z= 3...3!2!>3!-- YES
Different answers -- Insuff

(2) x + y = z
OR x!y!>(x+y)!
It will always be NO, since
x! = 1*2*3...(x-1)*x
y!= 1*2*3...*(y-1)*y
(x+y)!= 1*2*3*...X..y.....(x+y)
x!*y! gives us square of the lower integers, whereas (x+y)! gives us product of higher values..
Examples
x=1 y=2 and z=3..
1!2!>3!----NO
x=3,y=3 and z=6
3!3!>6!---NO
Suff

B
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Re: If x, y, and z are positive integers, is (x!)(y!) > z! ? [#permalink] New post  05 Apr 2016, 10:53
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x, y and z are positive integers. Is (x!)(y!) > z! ?

St1: x^y > z
2^1 > 1 --> Is (x!)(y!) > z! ? Yes
2^10 > 1000 --> Is (x!)(y!) > z! ? No
Not sufficient.

St2: x + y = z --> Is (x!)(y!) > (x + y)!
In this case, z! will have all the factors of x!, y! and (x + y)! We will have more multiplication operations on RHS and as a result LHS will always be lesser than RHS.
For ex: (2!)(3!) = (2*1)*(3*2*1) = 12
5! = 5*4*3*2*1 = 120
Hence (x!)(y!) < z!
Sufficient.

Answer: B
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Faiq87
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If x, y, and z are positive integers, is (x!)(y!) > z! ? [#permalink] New post  29 Jan 2017, 12:27
chetan2u wrote:
Bunuel wrote:
If x, y, and z are positive integers, is (x!)(y!) > z! ?

(1) x^y > z
(2) x + y = z



Hi,

lets see the choices and substitue values to see answer

(1) \(x^y > z\)
x = 3, y=2 and z=6... 3!2!>6!-- NO
x=3, y=2 and z= 3...3!2!>3!-- YES
Different answers -- Insuff

(2) x + y = z
OR x!y!>(x+y)!
It will always be NO, since
x! = 1*2*3...(x-1)*x
y!= 1*2*3...*(y-1)*y
(x+y)!= 1*2*3*...X..y.....(x+y)
x!*y! gives us square of the lower integers, whereas (x+y)! gives us product of higher values..
Examples
x=1 y=2 and z=3..
1!2!>3!----NO
x=3,y=3 and z=6
3!3!>6!---NO
Suff

B



Hi let us assume x=0 and y=2 then x+y=2 then z=2 too

doesnt it make 2nd statement insuff too?
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Re: If x, y, and z are positive integers, is (x!)(y!) > z! ? [#permalink] New post  27 Jun 2019, 07:57
Bunuel wrote:
If x, y, and z are positive integers, is (x!)(y!) > z! ?

(1) x^y > z
(2) x + y = z


B

Statement: (x!)(y!) > z!

#1
3^2 > 1, statement is TRUE
3^2 > 6, statement is FALSE

#2

For any value:
4+4=8,
1+7=8,
8! > 7!
8!> 4!*4!. Sufficient.


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Re: If x, y, and z are positive integers, is (x!)(y!) > z! ? [#permalink] New post  17 May 2021, 10:32
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Re: If x, y, and z are positive integers, is (x!)(y!) > z! ? [#permalink]
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