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If in the expansion of $(1 + x)^m \cdot (1 – x)^n $, the coefficients of $ x $ and $ x^2 $are 3 and -6 respectively, then m is ?

I solved it in the following way :

Expanding we get, the coefficient of $ x $ as $ (m-n) = 3 \cdots (1)$ and coefficient of $ x^2 $ as $ \frac{n(n-1)}{2} + \frac{m(m-1)}{2} +- m \cdot n = -6 \cdots (2)$ after substituting and some algebraic treatment I got m = 12 and n = 9.

Now this is correct but I am interested if there exists any short procedure such that the entire problem could be done under a mint.

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    The only thing that jumps out at me is the appearance of a convenient square trinomial: clearing the fractions and expanding your equation (2), we get $n^2 - n + m^2 - m - 2 m n = -12$. The second-degree terms combine nicely: $n^2+m^2-2mn = (n-m)^2$, and your equation (1) tells us we can replace this value with $3^2$. This leaves you with the system $m-n=3$ and $m+n=21$, which is very easy to solve.2010-10-26
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    I didn't understand what you meant , precisely `clearing the fractions and expanding your equation (2), we get n2−n+m2−m−2mn=−12.` ?!2010-10-26
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    If you write small enough, you could perhaps fit the solution under a mint. But I think you mean "under a minute". ;-)2010-10-26
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    @Tretwick I should probably have written "denominators", not "fractions". (Too late to edit my comment.) If you multiply each term of your equation (2) by 2, you get $n(n-1) + m(m-1) - 2mn = -12$; on the left-hand side, this has "cleared the denominators", which tends to make an equation like this easier to handle. (Fractions can be a little intimidating ... but now we don't have any!) By "expanding", I simply mean to multiply-out the terms $n(n-1)$ and $m(m-1)$, which gives the equation $n^2-n+m^2-m-2mn=-12$.2010-10-26
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    Instead of binomial formula you can also derivate and plugin $x=0$ ;)2011-12-30

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For the coefficient of $x$ to be 3, $m-n=3$ as you said, so $m>n$ and $m-n>0$ and we can factor the original product: $$\begin{align} (1+x)^m(1-x)^n&=(1+x)^{m-n+n}(1-x)^n \\ &=(1+x)^{m-n}(1+x)^n(1-x)^n \\ &=(1+x)^{m-n}\left((1+x)(1-x)\right)^n \\ &=(1+x)^{m-n}(1-x^2)^n \\ &=(1+x)^3(1-x^2)^n \end{align}$$ Now, the coefficient of $x^2$ is the sum of the coefficients of $x^2$ in $(1-x^2)^n=1-nx^2+\cdots$ and $(1+x)^3=1+3x+3x^2+x^3$, so $-6=-n+3$ and $n=9$. Since $m-n=3$, $m=12$.

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    Apologies! it was a typo :)2010-10-26
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    @Tretwick Marain: Ah. Seeing your edit above, I've removed that line from my answer.2010-10-26
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    but how you factor $(1 + x)^m \cdot (1 – x)^n$ to $(1-x^2)^n(1+x)^{m-n}$ ? My maths is not good enough so could you please elaborate a bit ?2010-10-26
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    @Tretwick Marain: I edited my answer to include more of an explanation of the factoring. Am I giving you enough information now?2010-10-26
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    I should probably note that the reason it occurred to me to try this is knowing that $(1+x)(1-x)=1-x^2$.2010-10-26
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    I considered a similar approach. However, starting with the observation that $m-n=3$ (which came from what amounts to some sophisticated expansion on Tretwick's part in his original solution), just so you can re-write the product and expand *again* seems like a lot of extra work to *force* $1-x^2$ to appear in the problem. If $1-x^2$ is the key here, I think there must be a more direct way to make use of it. (Then again, maybe $1+x$ and $1-x$ were chosen to appear the problem simply because their powers are easy to expand.)2010-10-26
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    @Day Late Don: I suppose my mind is stuck in contest-problem mode as that's been about 80% of my waking hours for the past several days, so that combined with him already having a straight-forward solution and wanting something elegant made me look for tricks. Seeing the difference of squares pattern, I jumped to wondering which was greater, m or n, and found that the linear coefficient, which was dictated by the easy-to-calculate linear terms in each expansion, gave m-n=3, and that made the refactoring all the prettier. Otherwise, I'd have done exactly what you suggested in your comment.2010-10-26
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    @Isaac: I wasn't criticizing your attempt; I was just commenting about what made me try something else. My first thought, too, was that $1+x$ and $1-x$ were *begging* to be multiplied together. In fact, I just made two more (unsuccessful) passes over the problem looking for a direct way to use their product in an elegant solution; the begging is so hard to ignore. I'm now wondering if it might be a deliberate distraction. :)2010-10-26
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    @Day Late Don: I wasn't taking it as criticism; I was just thinking aloud on how I ended up there (and taking a break from writing contest questions).2010-10-26