$\begin{array}{1 1}(A)\;35\\(B)\;37\\(C)\;39\\(D)\;41\end{array} $

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- $C(n,r)=\large\frac{n!}{r!(n-r)!}$

Total '-' signs =4

Total '+' signs =6

$-+-+-+-+-+-+-$

Hence possible place of '-' sign =7

$\therefore$ The total no of ways =$7C_4$

$\Rightarrow \large\frac{7!}{4!3!}$

$\Rightarrow \large\frac{7\times 6\times 5\times 4!}{4!\times 3\times 2}$

$\Rightarrow 35$

Total no of ways =35

Hence (A) is the correct answer.

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