$\begin{array}{1 1} 7!4! \\ ^8P_2 \\ ^8P_4 \\ ^8 C_4 \end{array}$

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Since no conditions are there for white balls,

All the 7 black balls can be arranged in one way.

White balls are to be placed in between the black balls.

There are 8 places for white balls and 4 white balls are there.

Any 4 places are to be selected out of 8 places in $^8C_4$ ways.

$\therefore $ The required no. of arrangements = $^8C_4\times 1$

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