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Re: y is greater equal 2 [#permalink]
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SusieSushi wrote:
Carcass wrote:
\(y \geq 2\)


Quantity A
Quantity B
\(\frac{y!}{5(y-2)!}\)
\(\frac{(y+1)!}{3(y-1)!}\)



A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.

Kudos for the right answer and explanation


If I were to plug in 2, I believe A becomes undefined whereas B becomes 1. In this particular case, would B be considered bigger? I am not sure how to compare a number against an undefined



If you mean (2-2) = 0, then 0! = 1.
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Re: y is greater equal 2 [#permalink]
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One divide option B by option A as the denominator never equals zero.

\(\frac{Option B}{ Option A} = \frac{5(y+1)!(y-2)!}{3y!(y-1)!} = \frac{5(y+1)}{3(y-1)} > 1\)
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Re: y is greater equal 2 [#permalink]
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First, simplify a little bit.

(y)! = y*(y-1)*(y-2)!
Quantity A is simplified to y*(y-1)/5

(y + 1)! = (y + 1)*y*(y - 1)!
Quantity B is simplified to y*(y+1)/3

Test for 2 (A = 0.4; B = 2) and 3 (A = 1.2; B = 4) to clearly see that for each y greater or equal 2 the quantity B is always greater.

Answer is B.
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Re: y is greater equal 2 [#permalink]
beckhamine wrote:
One divide option B by option A as the denominator never equals zero.

\(\frac{Option B}{ Option A} = \frac{5(y+1)!(y-2)!}{3y!(y-1)!} = \frac{5(y+1)}{3(y-1)} > 1\)


This is a SUPER clever way to solve this problem! +1
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Re: y is greater equal 2 [#permalink]
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Re: y is greater equal 2 [#permalink]
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