The number of ways in which `m+n(nlem+1)` different things can be arranged in a row such that no two of the n things may be together, is

Options

A. `((m+n)!)/(m!n!)`
B. `(m!(m+1)!)/((m+n)!)`
C. `(m!(m+1)!)/((m-n+1)!)`
D. none of these

Answer

To solve the problem of arranging \( m+n \) different things in a row such that no two of the \( n \) things are together, we can follow these steps:

### Step-by-Step Solution

1. **Arrange the \( m \) things**:
First, we arrange the \( m \) different things. The number of ways to arrange \( m \) things is given by \( m! \).

**Hint**: Remember that the arrangement of \( m \) distinct items is calculated using factorial notation.

2. **Identify the gaps for \( n \) things**:
After arranging the \( m \) things, there will be \( m + 1 \) gaps where the \( n \) things can be placed. This includes one gap before the first item, one gap after the last item, and \( m - 1 \) gaps between the \( m \) items.

**Hint**: Visualize the arrangement of \( m \) items to see where the gaps are located.

3. **Choose gaps for \( n \) things**:
We need to choose \( n \) gaps from the \( m + 1 \) available gaps to place the \( n \) things. The number of ways to choose \( n \) gaps from \( m + 1 \) is given by \( \binom{m+1}{n} \).

**Hint**: Use the combination formula \( \binom{n}{r} = \frac{n!}{r!(n-r)!} \) to calculate the number of ways to choose gaps.

4. **Arrange the \( n \) things**:
Once we have chosen the gaps, we can arrange the \( n \) things in those gaps. The number of ways to arrange \( n \) things is given by \( n! \).

**Hint**: Factorial notation applies here as well since the arrangement of distinct items is involved.

5. **Combine the results**:
The total number of arrangements is the product of the number of ways to arrange the \( m \) things, the number of ways to choose the gaps, and the number of ways to arrange the \( n \) things. Therefore, the total number of arrangements is:

\[ \text{Total arrangements} = m! \times \binom{m+1}{n} \times n! \]

6. **Final expression**:
Substituting the combination formula into the total arrangements gives:

\[ \text{Total arrangements} = m! \times \frac{(m+1)!}{n!(m+1-n)!} \times n! \]

Simplifying this, we find:

\[ \text{Total arrangements} = \frac{(m+1)!}{(m+1-n)!} \times m! \]

### Final Answer
Thus, the number of ways in which \( m+n \) different things can be arranged in a row such that no two of the \( n \) things are together is:

\[ \frac{(m+1)!}{(m+1-n)!} \times m! \]