How Many Ways Are There to Distribute 13 Identical Balls into 10 Distinct Boxes With Each Box Having Either 1, 2, or 3 Balls?

To solve the problem of distributing 13 identical balls into 10 distinct boxes with each box containing either 1, 2, or 3 balls, we can use methods such as generating functions or combinatorial counting. This article will explore both approaches and provide a detailed solution.

Problem Statement

We need to find the number of ways to distribute 13 identical balls into 10 distinct boxes such that each box contains either 1, 2, or 3 balls. Let ( x_i ) represent the number of balls in box ( i ), where ( i ) ranges from 1 to 10. The conditions given can be expressed as:

( x_1 x_2 cdots x_{10} 13 )

To simplify this problem, we can redefine the variables. Let:

( y_i x_i - 1 )

This transformation means that ( y_i ) can take values in ( {0, 1, 2} ) since ( x_i ) can take values 1, 2, or 3. The equation now becomes:

( y_1 1 y_2 1 cdots y_{10} 1 13 )

which simplifies to:

( y_1 y_2 cdots y_{10} 3 )

Now we need to count the number of non-negative integer solutions to the equation ( y_1 y_2 cdots y_{10} 3 ) with the additional constraint that ( y_i leq 2 ).

Solution Method

Total Solutions Without Constraints

First, we calculate the total number of solutions to ( y_1 y_2 cdots y_{10} 3 ) without considering the upper limit of 2 on each ( y_i ). This can be found using the binomial coefficient:

The number of solutions in non-negative integers to the equation ( y_1 y_2 cdots y_k n ) is given by:

( binom{n k - 1}{k - 1} )

In our case, ( n 3 ) and ( k 10 ):

( text{Total solutions} binom{3 10 - 1}{10 - 1} binom{12}{9} binom{12}{3} 220 )

Subtract the Invalid Solutions

Next, we need to subtract the cases where one or more ( y_i ) exceed 2. Suppose ( y_i ) exceeds 2. We can let ( y_i y_i - 3 ) which gives us:

( y_1 y_2 cdots y_{10} 0 )

This means ( y_1 ) is non-negative, and all other ( y_i ) are still non-negative. The number of solutions to this equation is given by:

( y_1 y_2 cdots y_{10} 0 )

This has only one solution, which is ( y_1 0 ) and ( y_2 0, y_3 0, ldots, y_{10} 0 ). This situation can occur for any of the 10 boxes. Therefore, we have:

( text{Invalid solutions} 10 times 1 10 )

Final Count of Valid Distributions

Now, we subtract the invalid solutions from the total solutions:

( text{Valid solutions} 220 - 10 210 )

Thus, the number of ways to distribute 13 identical balls into 10 distinct boxes with each box having either 1, 2, or 3 balls is:

( boxed{210} )

Conclusion

This problem can be solved using methods such as generating functions or combinatorial counting, demonstrating the power of these techniques in solving complex distribution problems. By transforming the problem and applying combinatorial principles, we can find the exact number of valid distributions, which in this case, is 210.