Question

The number of ways of arranging $m$ positive and $n(<m+1)$ negative signs in a row so that no two negative signs are together is

• A. ${}^{m+1}{P}_n$
• B. ${}^{n+1}{P}_m$
• C. ${}^{m+1}{C}_n$
• D. ${}^{n+1}{C}_m$

Answer 1

The correct option is C ${}^{m+1}{C}_n$

First arrange $m$ positive signs. The number of ways is just 1 (as all $+$ signs are identical).
Now, $m+1$ gaps are created of which n are to be selected for placing ${}^{\prime }{-}^{\prime }$ signs.
Then the total number of ways of doing so is ${}^{m+1}{C}_n$.
After selecting the gaps ${}^{\prime }{-}^{\prime }$ signs can be arranged in one way only.

Hence, there are total ${}^{m+1}{C}_n$ ways.