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I'm doing binomial expansion and I'm rather confused at how people can find a certain coefficient of certain rows.

For example, if a problem was $(2x - 10y)^{54}$, and I were to figure out the $32^{\text{nd}}$ element in that expansion, how would I figure out?

Would I have to look at or draw out a Pascal's triangle, then go 1 by 1 until I hit row 54? Is there an equation that would tell me what the xth element of the nth row is by plugging in numbers?

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The $n^{th}$ row reads

$$1,n,\frac{n(n-1)}2,\frac{n(n-1)(n-2)}{2\cdot3},\frac{n(n-1)(n-2)(n-3)}{2\cdot3\cdot4}\cdots$$

This is computed by recurrence very efficiently, like

$$1,54,\frac{54\cdot53}2=1431,\frac{1431\cdot52}3=24804,\frac{24804\cdot51}4=316251\cdots$$

Using symmetry, only the first half needs to be evaluated. Compared to the factorial formula, this is less prone to overflows.

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Hint: $(a+b)^n=\sum\limits_{k=0}^n {n\choose k }a^kb^{n-k}$ where ${n\choose k}=\frac{n!}{k!(n-k)!}$

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The nth row of a pascals triangle is:

$$_nC_0, _nC_1, _nC_2, ...$$

recall that the combination formula of $_nC_r$ is

$$ \frac{n!}{(n-r)!r!}$$

So element number x of the nth row of a pascals triangle could be expressed as

$$ \frac{n!}{(n-(x-1))!(x-1)!}$$

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  • $\begingroup$ Welcome to MSE. Your answer adds nothing new to the already existing answers. $\endgroup$ – José Carlos Santos Dec 30 '18 at 19:46

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