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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.