The number of ways of arranging m+ive signs n -ive signs (`n lt m +1`) in a row so that no two -ive signs are together is

Options

A. `""^(m+1)P_n`
B. `"^(m+1)C_n`
C. `"^(n+1)P_m`
D. `"^(n+1)C_m`

Answer

To find the number of ways to arrange \( m \) positive signs and \( n \) negative signs (where \( n < m + 1 \)) in a row such that no two negative signs are together, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Arrangement**:
We have \( m \) positive signs and \( n \) negative signs. The condition that no two negative signs can be together implies that we need to place the negative signs in such a way that they are separated by positive signs.

2. **Placing Positive Signs**:
First, we arrange the \( m \) positive signs in a row. This creates gaps where we can place the negative signs. For \( m \) positive signs, the arrangement would look like this:
\[ + + + + \ldots + \quad (m \text{ times}) \]

3. **Identifying Gaps**:
When we arrange \( m \) positive signs, there are \( m + 1 \) gaps created:
- One gap before the first positive sign
- One gap between each pair of positive signs
- One gap after the last positive sign

Thus, the total number of gaps is \( m + 1 \).

4. **Placing Negative Signs**:
We need to place \( n \) negative signs in these \( m + 1 \) gaps. The condition \( n < m + 1 \) ensures that we have enough gaps to place the negative signs without them being adjacent.

5. **Choosing Gaps for Negative Signs**:
We can choose \( n \) gaps from the \( m + 1 \) available gaps. The number of ways to choose \( n \) gaps from \( m + 1 \) is given by the combination formula:
\[ \binom{m + 1}{n} \]

6. **Final Answer**:
Therefore, the total number of ways to arrange \( m \) positive signs and \( n \) negative signs such that no two negative signs are together is:
\[ \boxed{\binom{m + 1}{n}} \]