You can sort of 'cheat' your way through this one.

Notice that both statements are 'symmetrical'. In other words, you could swap all of the As and Bs around, and neither statement would change. Therefore, if you could find two values for a and b that fit with one or both of the statements, you could also swap those values and use them.

Here's a simpler example of a symmetrical problem:



1 and 10 work with the first statement, so 10 and 1 will work as well. 3 and 4 work with the second statement, so 4 and 3 work as well. A symmetrical statement isn't going to ever be sufficient for this kind of question! (Where it asks you for the value of just one of the two variables.)

The only exception would be if a and b always had to be equal. That seems unlikely, since the second statements says that a!+b! = 3; half of 3 is 1.5, and factorials can't be decimals in GMAT land.

Since you could definitely find multiple values that would work for a by just swapping a and b backwards, the answer is going to have to be E. On test day, I'd want to take the time to write it out, but if I was rushing, this would be a great guess.
_________________

from 1
a*b*c=c!
means that c>b>a and a,b,c are all consective integers
so value of a at best can be 1 only
1*2*3=3! or 2*3*4=4!

not sufficient

From 2:
a!+b!=3
only possible if
a=1 and b=2 or a=2 ., b=2
Not sufficient

IMO E ..

IMO E

If a, b, and c are positive integers, what is the value of a?

(1) a*b*c = c!
(2) a! + b! = 3

a b c are positive integers ie they are intergers greater than zero

Stmt 1 : A*b*c=c!
1*2*3= 3! =6
also
2*3*4= 4! =24

so a can be 1 or 2 NS

Stmt 2: a!+b!=3

1+2=3
2+1=3
so again a can be 1 or 2 NS

Combining - same information
SO E

Par of GMAT CLUB'S New Year's Quantitative Challenge Set

_________________

Statement 1: In this expression, a, b, c can take up many values such as:

1*2*3=3!
or
2*3*4=4!
or
1*2*2=2! (Notice the a, b, c don't need to be distinct)
or even
1*1*1=1!

Thus, we can't come up with a unique value for a.

Thus, this statement is insufficient

Statement 2: Here, a! and b! can only occupy two values: 2 and 1. This is because 3 can be formed by only two combinations of two positive integers.

But, here too, we can't come up with a unique value for a

Thus, this statement is insufficient

Combining 1 & 2 we can come up with two values of a: 1 & 2

Thus, answer is E