Question

Find the total number of ways in which six ${}^{\prime }{+}^{\prime }$ and four ${}^{\prime }{-}^{\prime }$ signs can be arranged in a line such that no two ${}^{\prime }{-}^{\prime }$ signs occur together.

Answer 1

Let $x_1$ be the number of ${}^{\prime }{+}^{\prime }$ signs inserted before first ${}^{\prime }{-}^{\prime }$ sign

let $x_2,x_3,x_4$ be the number of ${}^{\prime }{+}^{\prime }$ signs inserted between ${}^{\prime }{-}^{\prime }$ signs

let $x_5$ be the number of ${}^{\prime }{+}^{\prime }$ signs inserted after last ${}^{\prime }{-}^{\prime }$ sign

$⟹x_1,x_5\ge 0,x_2,x_3,x_4\ge 1$

and also $x_1+x_2+x_3+x_4+x_5=6$

$x_1+(x_2-1)+(x_3-1)+(x_4-1)+x_5=3$

no of ways for this is ${}^{n+r-1}{C}_{r-1}$ where $n=3,r=5$

${=}^{5+3-1}{C}_{5-1}{=}^7{C}_4=35$