Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

Hi MathRevolution Firstly this Ain't a probability Question Kindly change that Here is my approach => (n+2)(n+1)*n!/n!=132=> (n+2)*(n+1)=132 Checking the values => n must be 10 => 12 *11=132

If (n+2)!/n!=132, n=? A. 2/131 B. 9 C. 10 D. 11 E. 12

*An answer will be posted in 2 days

(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:

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:

Once you get to (n+2)*(n+1)=132 (or further n² + 3n + 2 = 132)

It's important to develop a "feel" for such equations. I'd just immediately look at the biggest term n² , we need an n that's at least close to 132 when squared, so 10^2 = 100 seems like a good starting point. If you just mentally plug that into either of those equations, you would realized you've hit jackpot. So definitely no factorization of quadratic or going through all the options required.