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I thought it is pretty intuitive :

3!/4! = 1/4 , I can't imagine in any other case where the factorials would divide in such a manner that the remaining number in denominator is the highest leftover from the factorial. Think of it like this -> n! = n * (n-1)!
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devinawilliam83
If a, b and c are positive integers, what is the value of (a+b)(c+d)

(1) 4a!= b!
(2) 5c!=d!

1. b!/a! = 4. which means that b=4, and a must be 3. what about c and d?
2. d!/c! = 5. d=5 and c=4. what about a and b?

1+2
we have all the values.
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Just to throw a wrench in here, we do have to consider the possibility that one of the a,b,c,d could be 1 or 0. For instance, if one of the statements read:

(1) 6c!=d!, then (c,d) could be (5,6) or it could also be (0,3) or (1,3).

However, neither 4 or 5 are factorial numbers so we're safe. Always good to think of these cases though!
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Answer:C
If a, b and c are positive integers, what is the value of (a+b)(c+d)
(1) 4a!= b!
(2) 5c!=d!
1- 4x3!=4! ⇒ a=3 and b=4 -- c and d unknown -- NOT SUFFICIENT
2- 5x4!=5! ⇒ c=4 and d=5 -- a and b unknown -- NOT SUFFICIENT
Taking --1-- and --2-- together, we have all 4 variables -- SUFFICIENT
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speedilly
Just to throw a wrench in here, we do have to consider the possibility that one of the a,b,c,d could be 1 or 0. For instance, if one of the statements read:

(1) 6c!=d!, then (c,d) could be (5,6) or it could also be (0,3) or (1,3).

However, neither 4 or 5 are factorial numbers so we're safe. Always good to think of these cases though!


what do you mean by "neither 4 or 5 are factorial numbers", and how are we safe, if c and d were 1 and 3, then the whole answer would change no?
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speedilly
Just to throw a wrench in here, we do have to consider the possibility that one of the a,b,c,d could be 1 or 0. For instance, if one of the statements read:

(1) 6c!=d!, then (c,d) could be (5,6) or it could also be (0,3) or (1,3).

However, neither 4 or 5 are factorial numbers so we're safe. Always good to think of these cases though!

Hi there
Your assumption is not true as the stem has mentioned that a,b,c,d are positive integer and '0' is not a positive integer.

Posted from my mobile device
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awdxzs
speedilly
Just to throw a wrench in here, we do have to consider the possibility that one of the a,b,c,d could be 1 or 0. For instance, if one of the statements read:

(1) 6c!=d!, then (c,d) could be (5,6) or it could also be (0,3) or (1,3).

However, neither 4 or 5 are factorial numbers so we're safe. Always good to think of these cases though!

Hi there
Your assumption is not true as the stem has mentioned that a,b,c,d are positive integer and '0' is not a positive integer.

Posted from my mobile device

Still, it's possible that one of them could be 1. If Statement 2 was instead 6c! = d!, then (c,d) has two possibilities and therefore the answer would be E.
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Asked: If a, b and c are positive integers, what is the value of (a + b)(c + d)

(1) 4a! = b!
b = 4
a = 3
Since c & d are unknown
NOT SUFFICIENT

(2) 5c! = d!
d = 5
c = 4
Since a & b are unknown
NOT SUFFICIENT

(1) 4a! = b!
b = 4
a = 3
(2) 5c! = d!
d = 5
c = 4
(a + b)(c + d) = (3 + 4) (4 + 5) = 7*9 = 63
SUFFICIENT

IMO C
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