$X$ represents sample in frequency domain and $x$ represents samples in time domain.

NOTATION 1

$X(k) = \sum\limits_{n = 0}^{N - 1} {x\left[ n \right]e^{ - j\frac{{2\pi }}{N}nk} } $

$ x\left[ n \right] = \frac{1}{N}\sum\limits_{k = 0}^{N - 1} {X\left[ k \right]e^{j\frac{{2\pi }}{N}nk} } $

NOTATION 2

$ X\left[ k \right] = \frac{1}{N}\sum\limits_{k = 0}^{N - 1} {x\left[ n \right]e^{ - j\frac{{2\pi }}{N}nk} } $

$ x\left[ n \right] = \sum\limits_{k = 0}^{N - 1} {X\left[ k \right]e^{j\frac{{2\pi }}{N}nk} } $

NOTATION 3

$ X\left[ k \right] = \frac{1}{{\sqrt N }}\sum\limits_{k = 0}^{N - 1} {x\left[ n \right]e^{ - j\frac{{2\pi }}{N}nk} } $

$ x\left[ n \right] = \frac{1}{{\sqrt N }}\sum\limits_{k = 0}^{N - 1} {X\left[ k \right]e^{j\frac{{2\pi }}{N}nk} } $

NOTATION 4

$ X\left[ K \right] = \sum\limits_{n = 0}^{N - 1} {x\left[ n \right]e^{j\frac{{2\pi }}{N}nk} } $

$ x\left[ n \right] = \frac{1}{N}\sum\limits_{n = 0}^{N - 1} {X\left[ K \right]e^{ - j\frac{{2\pi }}{N}nk} } $

Please observe the scaling factor $\frac{1}{N}$ and change of a negative sign $-$ over the exponent term ${e^{\pm j\frac{{2\pi }}{N}nk} }$. Why is every notation of DFT valid?