In how many we can put 6 identical white balls and 5 identical black balls in 10 different boxes if each box must contain atleast one ball.

Answer

To solve the problem of distributing 6 identical white balls and 5 identical black balls into 10 different boxes with the condition that each box must contain at least one ball, we can follow these steps:

### Step 1: Understand the total number of balls
We have a total of 6 white balls and 5 black balls. Therefore, the total number of balls is:
\[ 6 + 5 = 11 \text{ balls} \]

### Step 2: Ensure each box has at least one ball
Since there are 10 boxes and each must contain at least one ball, we will first place one ball in each of the 10 boxes. This means we will use 10 balls out of the 11 we have.

### Step 3: Calculate remaining balls
After placing one ball in each of the 10 boxes, we have:
\[ 11 - 10 = 1 \text{ ball remaining} \]
This remaining ball can be either a white or a black ball.

### Step 4: Distributing the remaining ball
Now, we need to distribute the remaining ball into any of the 10 boxes. Since we have 10 boxes to choose from, we can place the remaining ball in any one of these boxes.

### Step 5: Calculate the number of ways to distribute the remaining ball
The number of ways to place the remaining ball into one of the 10 boxes is simply:
\[ 10 \text{ ways} \]

### Conclusion
Thus, the total number of ways to distribute the 6 identical white balls and 5 identical black balls into 10 different boxes, ensuring that each box has at least one ball, is:
\[ 10 \]