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­If x and y are positive integers, is x > y ?

(1) x! = 1

(2) y = 1­

in question we know both x and y >0
(1) x! = 1 its mean x <= y suff
(2) y=1 ,y=y or x>y insuff
hence (A)
How can it be concluded that if x!=1, then x must be less than equal to y, since the relation between x and y is neither established in the first statement nor in the beginning ? It's only known that x and y are both positive integers.

For the factorial of a non-negative integer to equal 1, the integer must be either 0 or 1. Only 0! = 1 and 1! = 1. Since it’s given that x is a positive integer, x! = 1 implies that x = 1. Since y is also a positive integer (1, 2, 3, and so on), the value of x, which is 1, cannot be greater than the value of y. Therefore, the answer to whether x > y is NO.
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in question we know both x and y >0
(1) x! = 1 its mean x <= y suff
(2) y=1 ,y=y or x>y insuff
hence (A)
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Remember only 0!=1
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Bunuel
Manushya
VLantoyMC
­If x and y are positive integers, is x > y ?

(1) x! = 1

(2) y = 1­

in question we know both x and y >0
(1) x! = 1 its mean x <= y suff
(2) y=1 ,y=y or x>y insuff
hence (A)
How can it be concluded that if x!=1, then x must be less than equal to y, since the relation between x and y is neither established in the first statement nor in the beginning ? It's only known that x and y are both positive integers.

For the factorial of a non-negative integer to equal 1, the integer must be either 0 or 1. Only 0! = 1 and 1! = 1. Since it’s given that x is a positive integer, x! = 1 implies that x = 1. Since y is also a positive integer (1, 2, 3, and so on), the value of x, which is 1, cannot be greater than the value of y. Therefore, the answer to whether x > y is NO.
And since factorial(x)=1, we can accordingly infer that x cannot be greater than y, it being non-negative anyway. Thanks for clarifying
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Remember only 0!=1

Right. But isn't 1! = 1x1 = 1 also since x and y are non-negative integers ?
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