There are five different boxes and seven different balls. The number of ways in which these balls can be distributed so that box 2 and box 4 contain only 1 ball each and at least 1 box is empty is N. (Order of putting the balls in the boxes is NOT considered). Then the digit in the hundredth's place of N is ............

Options

A. a. 1100
B. b. 2200
C. c. 3906
D. d. 4400

Answer

To solve the problem, we need to find the number of ways to distribute 7 different balls into 5 different boxes under the given conditions. Let's break it down step by step.

### Step 1: Choose the Balls for Box 2 and Box 4
We need to select 1 ball for Box 2 and 1 ball for Box 4. Since there are 7 different balls, we can choose the ball for Box 2 in \( \binom{7}{1} \) ways and the ball for Box 4 in \( \binom{6}{1} \) ways (after choosing the ball for Box 2, only 6 balls remain).

- **Calculations**:
\[ \text{Ways to choose for Box 2} = \binom{7}{1} = 7 \]
\[ \text{Ways to choose for Box 4} = \binom{6}{1} = 6 \]

### Step 2: Total Ways for Boxes 2 and 4
The total number of ways to choose 1 ball for Box 2 and 1 ball for Box 4 is:
\[ 7 \times 6 = 42 \]

### Step 3: Remaining Balls and Boxes
After placing 1 ball in Box 2 and 1 ball in Box 4, we have 5 balls left and 3 boxes remaining (Box 1, Box 3, and Box 5). We need to distribute these 5 balls into the 3 boxes such that at least one box remains empty.

### Step 4: Total Distributions Without Restrictions
First, we calculate the total number of ways to distribute 5 balls into 3 boxes without any restrictions. Each ball can go into any of the 3 boxes, so the total number of distributions is:
\[ 3^5 \]

### Step 5: Subtract the Cases Where No Box is Empty
To find the distributions where at least one box is empty, we can use the principle of inclusion-exclusion. We first calculate the total distributions where no box is empty using the formula for surjective functions (onto functions) which is given by:
\[ \text{Number of onto functions} = k! \cdot S(n, k) \]
where \( S(n, k) \) is the Stirling number of the second kind, representing the number of ways to partition \( n \) objects into \( k \) non-empty subsets.

For our case, \( n = 5 \) (balls) and \( k = 3 \) (boxes):
\[ \text{Number of onto functions} = 3! \cdot S(5, 3) \]
Calculating \( S(5, 3) \):
\[ S(5, 3) = 25 \quad (\text{known value}) \]
Thus,
\[ \text{Number of onto functions} = 6 \cdot 25 = 150 \]

### Step 6: Total Ways with At Least One Box Empty
Now, we can find the number of ways to distribute the 5 balls into 3 boxes with at least one box empty:
\[ \text{Total ways} = 3^5 - \text{Number of onto functions} = 243 - 150 = 93 \]

### Step 7: Final Calculation of N
Now, we multiply the number of ways to choose balls for Box 2 and Box 4 by the number of ways to distribute the remaining balls:
\[ N = 42 \times 93 = 3906 \]

### Step 8: Find the Digit in the Hundredth's Place
To find the digit in the hundredth's place of \( N \):
- The number \( 3906 \) has the hundredth's place digit as \( 3 \).

### Final Answer
The digit in the hundredth's place of \( N \) is **3**.

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