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If (n+2)!/n!= 132, n=? A. 2/131
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12 Oct 2021, 09:57
12 Oct 2021, 09:57
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Re: If (n+2)!/n!= 132, n=? A. 2/131
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12 Oct 2021, 10:05
12 Oct 2021, 10:05
Carcass wrote:
If \(\frac{(n+2)!}{n!}= 132\), n=?
A. 2/131
B. 9
C. 10
D. 11
E. 12
A. 2/131
B. 9
C. 10
D. 11
E. 12
(n+2)!/n! = 132
Rewrite as: [(n+2)(n+1)(n)(n-1)(n-2)....(3)(2)(1)]/[(n)(n-1)(n-2)....(3)(2)(1)] = 132
Cancel out terms: (n+2)(n+1) = 132
From here, we might just TEST the answer choices.
Since (12)(11) = 132, we can see that n = 10
Answer: C
Alternatively, we can take (n+2)(n+1) = 132 and expand it to get: n² + 3n + 2 = 132
Subtract 132 from both sides to get: n² + 3n - 130 = 0
Factor to get: (n + 13)(n - 10) = 0
So, n = -13 or n = 10
Since n cannot be negative in a factorial, n must equal 10
Answer: C
Cheers,
Brent
General Discussion
Retired Moderator
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Status:GRE Quant Tutor
Posts: 1141
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Concentration: General Management
Schools: XLRI Jamshedpur, India - Class of 2014
GMAT 1: 700 Q51 V31
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WE:Engineering (Computer Software)
Re: If (n+2)!/n!= 132, n=? A. 2/131
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26 Jul 2025, 01:27
26 Jul 2025, 01:27
1
\(\frac{(n+2)!}{n!}= 132\)
=> \(\frac{(n+2)*(n+1)*n!}{n!}= 132\)
=> (n+1)*(n+2) = 132 = 11*12 = (10+1)*(10+2)
=> n = 10
So, Answer will be C
Hope it helps!
Playlist on Solved Problems on Sequences here
=> \(\frac{(n+2)*(n+1)*n!}{n!}= 132\)
=> (n+1)*(n+2) = 132 = 11*12 = (10+1)*(10+2)
=> n = 10
So, Answer will be C
Hope it helps!
Playlist on Solved Problems on Sequences here
