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Kudos please if explanation helped ------------------------------------------------------------------------------------------------- Don't stop when you are tired , stop when you are DONE .

The question tells us that (n + 2)! = n!(an² + bn + c) So, we can write: (n!)(n + 2)(n + 1) = n!(an² + bn + c) Divide both sides by n! to get: (n + 2)(n + 1) = an² + bn + c Use FOIL to expand left side: n² + 3n + 2 = an² + bn + c In other words: 1n² + 3n + 2 = an² + bn + c

So, a = 1, b = 3 and c = 2 This means abc = (1)(3)(2) = 6

I was wondering could you please explain where I am making a mistake in my solution? I am getting different answers.

Let's say n=1

6=1(1a+1b+c). If we were to pick a=b=1 and c=4, abc=6 BUT if we were to pick a=b=c=2, we get 8 instead.

Can we not pick smart numbers for this question specifically?

If you want to solve by plugging numbers here, you should look at for possible answer but then test these numbers in other values of n.. for example take n as 2 now.. \((2+2)!=2!(a2^2+2b+c).......24=2(4a+2b+c)..........12=4a+2b+c\) the values of (a, b, c) when n=1 were (1, 1, 4) or (1,4,1) or (4,1,1) or (2,2,2) or (1,2,3) or (1,3,2) and so on.. Only (1,3,2) satisfies the equation when n=2.

So, the method of plugging would become a bit more cumbersome, although it will give you a correc answer after few steps
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