12!/(3^4*5!*2^6)=

\(\frac{12!}{(3^4*5!*2^6)}=\)

\(=\frac{12*11*10*9*8*7*6*5*4*3*2}{(3*3*3*3)*(5*4*3*2*1)*(2*2*2*2*2*2)}=\)

\(=\frac{(2*2*3)*11*(2*5)*(3*3)*(2*2*2)*7*(2*3)*5*(2*2)*3*2}{(3*3*3*3)*(5*(2*2)*3*2)*(2*2*2*2*2*2)}=\)

\(=\frac{2^{10}*3^5*11*5^2*7}{3^5*5*2^9}=\)

\(=2*11*5*7=2*11*35=2*385=770=B\)

MANHATTAN GMAT OFFICIAL SOLUTION:

Since 5! and 12! share the common terms (5)(4)(3)(2)(1), these terms can be canceled from both the numerator and denominator. Thus, we can rewrite \(\frac{12!}{(3^4*5!*2^6)}\) as:

\(\frac{12*10*9*8*7*6}{(3^4*2^6)}=\)

We need to cancel the 2's and 3's in the denominator, so we identify these common factors in the numerator.

\(\frac{(2^2*3)*11*(2*5)*(3^2)*(2^3)*7*(2*3)}{(3^4*2^6)}\)

Group the common factors of 2 and 3 in the numerator, then cancel and compute:

\(\frac{(2^7*3^4)*11*5*7}{(3^4*2^6)}=2*11*5*7=770\)

The correct answer is B.

Note: If you do not have time to do all the canceling involved in this problem, you can at least rule out answer choices C (480) and E (35), because the correct answer must be a multiple of 11. The 12! in the numerator is a multiple of 11, and the denominator contains no factors of 11 that would cancel this term. You could use the same argument to eliminate non-multiples of 7; however, the divisibility rule for 7 is complicated, and on typical GMAT problems, testing by long division or inspection will be just as fast. In any event, you could eliminate answer choices C (480) and A (2,210) in this manner.

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