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# In how many ways can 7 plus (+) signs and 4 minus (-) signs can be arranged in a row so that no two minus signs are together?

$\begin{array}{1 1}(11)! \\ ^7C_4 \\ 7!.4! \\ 70 \end{array}$

$i.e.,$ there are 8 places for minus (-) and 4 (-) signs are there.
$\therefore$ these minus (-) signs can be placed in $^8C_4$ ways.
The required no. of arrangements = $^8C_4=70$