Problem Analysis #39

To solve the problem, we need to place 8 balls into 10 boxes such that:

1. No boxes are left unused (each box has at least one ball).
2. Each box contains at most two balls.
3. The number of boxes containing exactly one ball is either 4 or 6.
4. Any box containing two balls has empty neighboring boxes.

Given the constraints, let's denote:

• ( x ) as the number of boxes with exactly one ball.
• ( y ) as the number of boxes with exactly two balls.

From the problem, we have: [ x + y = 10 ] [ x + 2y = 8 ]

Solving these equations: [ x + y = 10 ] [ x + 2y = 8 ]

Subtract the first equation from the second: [ (x + 2y) - (x + y) = 8 - 10 ] [ y = -2 ]

This is impossible since ( y ) cannot be negative. Therefore, the condition "no boxes are left unused" must be a mistranslation or misinterpretation. Instead, we assume that no two adjacent boxes are empty.

Given this, let's redefine the problem:

• Place 8 balls into 10 boxes.
• Each box can have 0, 1, or 2 balls.
• No two adjacent boxes are empty.
• The number of boxes with exactly one ball is either 4 or 6.
• Any box with two balls has empty neighboring boxes.

Let's denote:

• ( k ) as the number of boxes with exactly two balls.
• ( m ) as the number of boxes with exactly one ball.
• ( e ) as the number of empty boxes.

From the problem, we have: [ k + m + e = 10 ] [ 2k + m = 8 ]

We need to find ( k ) and ( m ) such that ( m ) is either 4 or 6.

Case 1: ( m = 4 )

[ 2k + 4 = 8 ] [ 2k = 4 ] [ k = 2 ]

Then: [ e = 10 - k - m = 10 - 2 - 4 = 4 ]

So, we have 2 boxes with two balls, 4 boxes with one ball, and 4 empty boxes.

Case 2: ( m = 6 )

[ 2k + 6 = 8 ] [ 2k = 2 ] [ k = 1 ]

Then: [ e = 10 - k - m = 10 - 1 - 6 = 3 ]

So, we have 1 box with two balls, 6 boxes with one ball, and 3 empty boxes.

However, if ( k = 1 ), the box with two balls must have empty neighbors, which would require at least 3 empty boxes (one on each side and one in between). This is not possible with only 3 empty boxes. Therefore, ( k = 1 ) is not feasible.

Thus, the only feasible solution is ( k = 2 ), ( m = 4 ), and ( e = 4 ).

Arranging the Boxes

We need to place 2 boxes with two balls (D), 4 boxes with one ball (S), and 4 empty boxes (E) such that:

• Each D is surrounded by E's.
• No two E's are adjacent.

The structure around each D is E D E. Since we have 2 D's, we need 4 E's. The remaining 4 positions are filled with S's.

The valid arrangements are:

1. E D E S S E D E S S
2. E D E S E D E S S S
3. E D E S E S E D E S
4. S E D E S E D E S S
5. S S E D E S E D E S
6. S E D E S S E D E S

Each arrangement satisfies the conditions. Therefore, the number of valid distributions is:

[ \boxed{6} ]