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# There are 7 identical black balls and 4 identical white balls. In how many ways can they be arranged in a row so that no two white balls are adjacent?

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

Any 4 places are to be selected out of 8 places in $^8C_4$ ways.
$\therefore$ The required no. of arrangements = $^8C_4\times 1$