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If r is an integer, and 700 < 10!/(10-r)!<1,000, then r=

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If r is an integer, and 700 < 10!/(10-r)!<1,000, then r= [#permalink]

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New post 21 Nov 2017, 18:23
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[GMAT math practice question]

If r is an integer, and \(700 < \frac{10!}{(10-r)!}<1,000\), then r=

A. 1
B. 2
C. 3
D. 4
E. 5
Spoiler: OA

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Re: If r is an integer, and 700 < 10!/(10-r)!<1,000, then r= [#permalink]

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New post 21 Nov 2017, 19:37
MathRevolution wrote:
[GMAT math practice question]

If r is an integer, and \(700 < \frac{10!}{(10-r)!}<1,000\), then r=

A. 1
B. 2
C. 3
D. 4
E. 5



Hi....

\(\frac{10!}{(10-r)!}=10*9*8..*(10-(r-1)\)
Now it is given this value is more than 700 and less than 1000
10*9*8=720
So 10-(r-1)=8.....r-1=10-8=2....r=2+1=3
C
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Re: If r is an integer, and 700 < 10!/(10-r)!<1,000, then r= [#permalink]

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New post 22 Nov 2017, 05:27
MathRevolution wrote:
[GMAT math practice question]

If r is an integer, and \(700 < \frac{10!}{(10-r)!}<1,000\), then r=

A. 1
B. 2
C. 3
D. 4
E. 5


An easy way to solve this question is to plug in the answers given, calculate the value of \(\frac{10!}{(10-r)!}\) and see whether it satisfies or violates the inequality in the problem prompt.

A/\(r = 1\)
This gives us: \(\frac{10!}{(10-r)!} = \frac{10!}{9!} = 10\). This violates the inequality given in the problem prompt.
B/ \(r = 2\)
This gives us: \(\frac{10!}{(10-r)!} = \frac{10!}{8!} = 10*9 = 90\). This also violates the inequality given in the problem prompt.
C/ \(r = 3\)
This gives us: \(\frac{10!}{(10-r)!} = \frac{10!}{7!} = 10*9*8 = 720\). This satisfies the inequality given in the problem prompt, therefore, this is our answer (we can afford to stop here since this is not a DS question nor is the unknown a polynomial/non-integer with multiple potential values).

Thus, the answer is C.
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If r is an integer, and 700 < 10!/(10-r)!<1,000, then r= [#permalink]

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New post 22 Nov 2017, 19:54
=>

Note that
\(\frac{10!}{(10-r)!} = 10 * (10-1) * … * (10-r+1)\).

Test different values of \(r\):
\(r=1 : 10\)
\(r=2 : 10 * 9 = 90\)
\(r=3 : 10 * 9 * 8 = 720\)
\(r=4 : 10 * 9 * 8 * 7 = 5040\)

Therefore,\(r = 3\).

The answer is C.

Answer: C
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Re: If r is an integer, and 700 < 10!/(10-r)!<1,000, then r= [#permalink]

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New post 06 Feb 2018, 11:55
chetan2u wrote:
MathRevolution wrote:
[GMAT math practice question]

If r is an integer, and \(700 < \frac{10!}{(10-r)!}<1,000\), then r=

A. 1
B. 2
C. 3
D. 4
E. 5



Hi....

\(\frac{10!}{(10-r)!}=10*9*8..*(10-(r-1))\)
Now it is given this value is more than 700 and less than 1000
10*9*8=720
So 10-(r-1)=8.....r-1=10-8=2....r=2+1=3
C



Hi Chetan,

Can you please explain how did you derive \(\frac{10!}{(10-r)!}=10*9*8..*(10-(r-1)\)
Re: If r is an integer, and 700 < 10!/(10-r)!<1,000, then r=   [#permalink] 06 Feb 2018, 11:55
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