Browse Questions

# The total number of ways in which six '+' and four '-' signs can be arranged in a line such that no two signs '-' occur together is

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

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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.