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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?

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

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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 = $^8C_4\times 1$
answered Sep 5, 2013 by rvidyagovindarajan_1

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