Question

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 is ______.

Answer 1

${}^{\prime }{+}^{\prime }$ signs can be put in a row in one way creating seven gaps shown as arrows:
Now $4^{\prime }{-}^{\prime }$ signs must be kept in these gaps. So, no two ${}^{\prime }{-}^{\prime }$ signs should be together.
Out of these $7$ gaps $4$ can be chosen in ${}^7{C}_4$ ways. Hence, required number of arrangements is
${}^7{C}_4{=}^7{C}_3=\frac{7\times 6\times 5}{3\times 2\times 1}=35$.