(x+2y)^10
My answer: cooficient of X^4=210* 3^6=153090
This cannot be right! Please help
I still do not understand why it is
$\displaystyle {10\choose 4} x^4 (2y)^6$, doesn't it have to be $\displaystyle {10\choose 6} x^4 (2y)^6$. The formula that my textbook gives is
$\displaystyle {n\choose n-j} x^j a^(n-j)$,
so in my example n=10, j=4
Please help to understand.
Thank you
That is easy: $\displaystyle {{N} \choose {k}} = {{N} \choose {N - k}}$
$\displaystyle {{10} \choose {4}} = {{10} \choose {10 - 4}}$
$\displaystyle \frac{{N!}}{{\left( {K!} \right)\left( {N - K} \right)!}} = \frac{{N!}}{{\left[ {N - K} \right]!\left( {N - \left( {N - K} \right)} \right)!}}
$
Correct, but I personally think it is best to use the iteration you are on. For example, 0, then 1 then 2 then 3 then 4 then 5, 6, 7, 8, 9, 10
This helps you keep track of where you are, you could technically do
10, 9, 8, 7, 6, 5, 6, 7, 8, 9, 10 or 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 0 or 0, 1, 2, 3, 4, 5, 4, 3, 2, 1, 0
Or heck, any combination in between, such as 10, 1, 8, 3, 6, 5, 6, 7, 2, 1, 10
But with all those terms, organization will be helpful, and doing it in order will probably keep you from making errors.
Also, don't forget to multiply that combination of (n, n-k) by the coefficient of x to the power of n- k, and y to the power of k.
so
$\displaystyle {{10} \choose {4}} (1x)^4(2y)^6$
$\displaystyle {{10} \choose {4}} (1^4 x^4) (2^6 y^6)$
$\displaystyle {{10} \choose {4}} 2^6 x^4 y^6$
$\displaystyle (210) (64) x^4 y^6$
$\displaystyle 13440 x^4 y^6$
On a side note, $\displaystyle {{10} \choose {6}} ={{10} \choose {4}}$ is the reason why pascal's triangle is symmetrical. (ie: coefficients of 1, 2, 1 and 1, 3, 3, 1 etc)
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So, for the purposes of visualizing what's going on, the total breakdown will be:
$\displaystyle {{10} \choose {0}} (x)^{10}(2y)^0 + {{10} \choose {1}} (x)^9(2y)^1 + {{10} \choose {2}} (x)^8(2y)^2 + {{10} \choose {3}} (x)^7(2y)^3 + {{10} \choose {4}} (x)^6(2y)^4$ $\displaystyle + {{10} \choose {5}} (x)^5(2y)^5 +$ $\displaystyle {{10} \choose {6}} (x)^4(2y)^6 + {{10} \choose {7}} (x)^3(2y)^7 + {{10} \choose {8}} (x)^2(2y)^8 + {{10} \choose {9}} (x)^1(2y)^9 + {{10} \choose {10}} (x)^0(2y)^{10}$
Notice how easy it is to find any given term, because the combinations are sequential you can just find $\displaystyle {{10}\choose{6}}$ to find our term. This would be a bit harrier if we randomly alternated between our 2 options, you could easily lose track of where you are in the middle, and would have to count from the beginning. That is why I recommend you count sequentially from 0 to n in your combinations.