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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?

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

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Since there is no condition for plus (+) sign, fix them in a row.
+ + + + + + +.
There are 6 places in between each plus and one before and one after these plus (+) signs.
$i.e.,$ 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 = $^8C_4=70$


answered Aug 20, 2013 by rvidyagovindarajan_1
edited Dec 20, 2013 by meenakshi.p

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