There are 5 different boxes and 7 different balls. All the balls are to be distributed in the 5 boxes placd in a row so that any box can receive any number of balls.
In how many ways can these balls be distributed so that box 2 and box 4 contain only 1 and 2 balls respectively?

Options

A. 5522
B. 8505
C. 2305
D. none of these

Answer

To solve the problem of distributing 7 different balls into 5 different boxes, with the condition that box 2 contains exactly 1 ball and box 4 contains exactly 2 balls, we can follow these steps:

### Step 1: Choose the balls for box 2
Since box 2 must contain exactly 1 ball, we need to choose 1 ball from the 7 available balls. The number of ways to choose 1 ball from 7 is given by the combination formula:

\[ \text{Ways to choose 1 ball} = \binom{7}{1} = 7 \]

### Step 2: Choose the balls for box 4
After placing 1 ball in box 2, we have 6 balls remaining. Box 4 must contain exactly 2 balls. The number of ways to choose 2 balls from the remaining 6 balls is given by:

\[ \text{Ways to choose 2 balls} = \binom{6}{2} = 15 \]

### Step 3: Distribute the remaining balls
After placing 1 ball in box 2 and 2 balls in box 4, we have 4 balls left (7 - 1 - 2 = 4). These 4 balls can be placed in any of the 5 boxes (including box 1, box 3, and box 5). Since each ball can go into any of the 5 boxes, the number of ways to distribute these 4 balls is:

\[ \text{Ways to distribute 4 balls} = 5^4 = 625 \]

### Step 4: Calculate the total number of distributions
To find the total number of ways to distribute the balls under the given conditions, we multiply the number of ways to choose the balls for box 2, the number of ways to choose the balls for box 4, and the number of ways to distribute the remaining balls:

\[ \text{Total ways} = \text{Ways to choose for box 2} \times \text{Ways to choose for box 4} \times \text{Ways to distribute remaining balls} \]

Substituting the values we calculated:

\[ \text{Total ways} = 7 \times 15 \times 625 \]

Calculating this gives:

\[ \text{Total ways} = 7 \times 15 = 105 \]
\[ 105 \times 625 = 65625 \]

Thus, the total number of ways to distribute the balls is **65625**.

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