The numberof ways in which 2n distinct letters (addressed) can be distributed in N distinct mail boxes such that there are exactly K letters `(n lt K le 2n)` in one of the mail boxes is

Options

A. `""^(2n)C_(K)`
B. `""^(2n)C_(K).N(N-1)^(2n-K)`
C. `""^(2n)C_(K).(N-1)^(2n-K)`
D. `""^(2n)C_(K)(2n-K)^(N-1.N`

Answer

To solve the problem of distributing \(2n\) distinct letters into \(N\) distinct mailboxes such that exactly \(K\) letters are in one specific mailbox, we can follow these steps:

### Step 1: Choose the Mailbox
We first choose which mailbox will contain exactly \(K\) letters. Since there are \(N\) distinct mailboxes, we have \(N\) choices for this step.

**Hint:** Think about how many options you have for selecting one mailbox from \(N\) available mailboxes.

### Step 2: Choose the Letters for the Mailbox
Next, we need to choose \(K\) letters from the \(2n\) distinct letters to place in the selected mailbox. The number of ways to choose \(K\) letters from \(2n\) letters is given by the binomial coefficient \(\binom{2n}{K}\).

**Hint:** Recall the formula for combinations, which helps in selecting \(K\) items from a total of \(2n\) items.

### Step 3: Distribute the Remaining Letters
After placing \(K\) letters in the chosen mailbox, we have \(2n - K\) letters left. These remaining letters can be distributed among the \(N - 1\) remaining mailboxes. Each of these \(2n - K\) letters can go into any of the \(N - 1\) mailboxes. Therefore, the number of ways to distribute these letters is \((N - 1)^{(2n - K)}\).

**Hint:** Consider how many choices each of the remaining letters has for placement in the remaining mailboxes.

### Step 4: Combine the Results
Now, we combine all the choices we made:
1. Choose the mailbox: \(N\)
2. Choose \(K\) letters for that mailbox: \(\binom{2n}{K}\)
3. Distribute the remaining letters: \((N - 1)^{(2n - K)}\)

Thus, the total number of ways to distribute the letters is:
\[ N \cdot \binom{2n}{K} \cdot (N - 1)^{(2n - K)} \]

### Final Answer
The number of ways in which \(2n\) distinct letters can be distributed in \(N\) distinct mailboxes such that there are exactly \(K\) letters in one of the mailboxes is:
\[ N \cdot \binom{2n}{K} \cdot (N - 1)^{(2n - K)} \]