Is there an equation that represents the nth row in Pascal's triangle?
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.
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
$$ \frac{n!}{(n-(x-1))!(x-1)!}$$