If m rupee coins and n ten paise coins are placed in a line, then the probability that the extreme coins are ten paise coins, is

Options

A. `""^( m + n) C_(m)`
B. `(n(n - 1))/((m + n) ( m + n - 1))`
C. `""^(m+n)P_(m)`
D. `""^(m + n)P_(n)`

Answer

To find the probability that the extreme coins are ten paise coins when we have \( m \) rupee coins and \( n \) ten paise coins arranged in a line, we can follow these steps:

### Step 1: Calculate the Total Outcomes
The total number of ways to arrange \( m \) rupee coins and \( n \) ten paise coins in a line is given by the formula for permutations of multiset:

\[ \text{Total Outcomes} = \frac{(m+n)!}{m! \cdot n!} \]

**Hint:** Remember that when arranging items of different types, we divide by the factorial of the counts of each type to account for indistinguishable items.

### Step 2: Calculate the Expected Outcomes
To find the expected outcomes where both extreme coins are ten paise coins, we fix the two extreme positions with ten paise coins. This leaves us with \( m+n-2 \) positions to fill with the remaining coins.

- The number of ten paise coins left is \( n-2 \) (since 2 are used for the extremes).
- The number of rupee coins remains \( m \).

Thus, the number of ways to arrange the remaining coins is:

\[ \text{Expected Outcomes} = \frac{(m+n-2)!}{m! \cdot (n-2)!} \]

**Hint:** Fixing positions reduces the total number of coins to arrange, so adjust your counts accordingly.

### Step 3: Calculate the Probability
The probability \( P \) that the extreme coins are ten paise coins can be calculated by taking the ratio of the expected outcomes to the total outcomes:

\[ P = \frac{\text{Expected Outcomes}}{\text{Total Outcomes}} = \frac{\frac{(m+n-2)!}{m! \cdot (n-2)!}}{\frac{(m+n)!}{m! \cdot n!}} \]

### Step 4: Simplify the Probability Expression
Now, simplify the expression:

\[ P = \frac{(m+n-2)! \cdot n!}{(n-2)! \cdot (m+n)!} \]

We can express \( (m+n)! \) as:

\[ (m+n)! = (m+n)(m+n-1)(m+n-2)! \]

Substituting this into our probability expression gives:

\[ P = \frac{(m+n-2)! \cdot n!}{(n-2)! \cdot (m+n)(m+n-1)(m+n-2)!} \]

Cancelling \( (m+n-2)! \):

\[ P = \frac{n!}{(n-2)! \cdot (m+n)(m+n-1)} \]

Now, \( n!/(n-2)! = n(n-1) \):

\[ P = \frac{n(n-1)}{(m+n)(m+n-1)} \]

### Final Answer
Thus, the probability that the extreme coins are ten paise coins is:

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