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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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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
  • $\begingroup$ Very clear answer, thank you; exactly what I needed to know. $\endgroup$ – sil Oct 15 at 9:03
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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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Looking at the first few lines of the triangle you will see that they are powers of 11 ie the 3rd line (121) can be expressed as 11 to the power of 2. This works till the 5th line which is 11 to the power of 4 (14641).

An equation to determine what the nth line of Pascal's triangle could therefore be n = 11 to the power of n-1

This works till you get to the 6th line. Using the above formula you would get 161051. The 6th line of the triangle is 1 5 10 10 5 1. Both numbers are the same. By inspection you will see that 161051 expressed in base 11 is in fact 1 5 10 10 5 1.

The equation could therefore be refined as:

n (base 11) = 11 to the power of n-1

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