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

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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 = $^8C_4=70$

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