The problem asks how many ordered quadruplets $(a, b, c, d)$ exist such that the sum of four whole numbers $a$, $b$, $c$, and $d$ is 15. Each of these numbers is a whole number, meaning $a, b, c,$ and $d$ are non-negative integers.

This is a combinatorial problem that can be solved using the "stars and bars" theorem. Here's the solution:

Step-by-Step Solution

1. Define the Equation: We need to find the number of solutions to $a + b + c + d = 15$ where $a$, $b$, $c$, and $d$ are whole numbers (non-negative integers).

2. Apply the Stars and Bars Method: The number of non-negative integer solutions to an equation of the form $x_1 + x_2 + \cdots + x_k = n$ is given by the formula $\binom{n + k - 1}{k - 1}$ where $k$ is the number of variables, and $n$ is the sum.

3. Substitute Values:

   • Here, $k = 4$ (for $a$, $b$, $c$, and $d$).
   • $n = 15$ (the total sum).

   So we need to calculate $\binom{15 + 4 - 1}{4 - 1} = \binom{18}{3}$

4. Calculate $\binom{18}{3}$: $\binom{18}{3} = \frac{18 \times 17 \times 16}{3 \times 2 \times 1} = \frac{4896}{6} = 816$

Conclusion

There are 816 ordered quadruplets $(a, b, c, d)$ such that $a + b + c + d = 15$.

Would you like a deeper explanation of any steps or further assistance?

Here are five related questions to explore further:

1. How would the solution change if the numbers $a$, $b$, $c$, and $d$ had to be positive integers?
2. How many ordered triples $(a, b, c)$ satisfy $a + b + c = 10$?
3. What is the stars and bars theorem, and how is it applied in combinatorics?
4. How would you solve for ordered quadruplets if the sum were changed to a number other than 15?
5. What would be the solution if $a$, $b$, $c$, and $d$ had an additional constraint, like $a \leq 5$?

Tip: For problems involving the number of ways to achieve a sum with non-negative integers, the stars and bars method is often effective for simplifying calculations.