Key Concepts and Formulas

• Generating Functions: A generating function represents a sequence of numbers as the coefficients of a power series. For a single die roll, the generating function is $G(x) = x + x^2 + x^3 + x^4 + x^5 + x^6$.
• Binomial Theorem: $(1+x)^n = \sum_{k=0}^n \binom{n}{k} x^k$, where $\binom{n}{k} = \frac{n!}{k!(n-k)!}$. For negative exponents: $(1-x)^{-n} = \sum_{k=0}^\infty \binom{n+k-1}{k} x^k$.
• Stars and Bars (Inclusion-Exclusion): A method to count integer solutions to equations with constraints.

Step-by-Step Solution

Step 1: Formulate the Generating Function for Four Die Rolls

We want to find the number of ways to get a sum of 16 when rolling a six-sided die four times. The generating function for a single die roll is $G(x) = x + x^2 + x^3 + x^4 + x^5 + x^6$. The generating function for four die rolls is the product of the generating functions for each roll. Therefore: $G_4(x) = (G(x))^4 = (x + x^2 + x^3 + x^4 + x^5 + x^6)^4$ We can rewrite this as: $G_4(x) = \left(x(1 + x + x^2 + x^3 + x^4 + x^5)\right)^4 = x^4 (1 + x + x^2 + x^3 + x^4 + x^5)^4$ Using the geometric series formula, $1 + x + x^2 + ... + x^n = \frac{1-x^{n+1}}{1-x}$, we get: $G_4(x) = x^4 \left(\frac{1-x^6}{1-x}\right)^4 = x^4 (1-x^6)^4 (1-x)^{-4}$

Step 2: Identify the Desired Coefficient

We need to find the coefficient of $x^{16}$ in $G_4(x)$. This represents the number of ways to get a sum of 16. Therefore we need to find $[x^{16}] G_4(x)$. $[x^{16}] \left(x^4 (1-x^6)^4 (1-x)^{-4}\right)$ We can factor out $x^4$ to simplify the expression: $[x^{16}] \left(x^4 (1-x^6)^4 (1-x)^{-4}\right) = [x^{12}] \left((1-x^6)^4 (1-x)^{-4}\right)$

Step 3: Expand the Binomial Terms

We need to expand $(1-x^6)^4$ and $(1-x)^{-4}$ using the binomial theorem.

For $(1-x^6)^4$, we use the binomial expansion $(a+b)^n = \sum_{k=0}^n \binom{n}{k} a^{n-k} b^k$: $(1-x^6)^4 = \binom{4}{0}(1)^4(-x^6)^0 + \binom{4}{1}(1)^3(-x^6)^1 + \binom{4}{2}(1)^2(-x^6)^2 + \binom{4}{3}(1)^1(-x^6)^3 + \binom{4}{4}(1)^0(-x^6)^4$ $(1-x^6)^4 = 1 - 4x^6 + 6x^{12} - 4x^{18} + x^{24}$ Since we are looking for terms that contribute to $x^{12}$ in the final product, we only need to consider terms up to $x^{12}$ from this expansion: $1 - 4x^6 + 6x^{12}$.

For $(1-x)^{-4}$, we use the generalized binomial theorem, which states that $(1-z)^{-n} = \sum_{k=0}^\infty \binom{n+k-1}{k} z^k$. Here, $z=x$ and $n=4$: $(1-x)^{-4} = \sum_{k=0}^\infty \binom{4+k-1}{k} x^k = \sum_{k=0}^\infty \binom{k+3}{k} x^k = \sum_{k=0}^\infty \binom{k+3}{3} x^k$ This expansion generates an infinite series of terms.

Step 4: Multiply the Expansions and Collect the Coefficient of $x^{12}$

Now we multiply the relevant parts of the two expansions: $[x^{12}] \left( (1 - 4x^6 + 6x^{12} - \dots) \times \left( \binom{3}{3}x^0 + \binom{4}{3}x^1 + \dots + \binom{15}{3}x^{12} + \dots \right) \right)$ To find the coefficient of $x^{12}$, we consider all pairs of terms (one from each expansion) whose powers of $x$ sum to 12:

1. Constant term from $(1-x^6)^4$ multiplied by $x^{12}$ term from $(1-x)^{-4}$: $(1) \times \left( \binom{12+3}{3} x^{12} \right) = \binom{15}{3} = \frac{15 \times 14 \times 13}{3 \times 2 \times 1} = 5 \times 7 \times 13 = 455$

2. Term with $x^6$ from $(1-x^6)^4$ multiplied by $x^6$ term from $(1-x)^{-4}$: $(-4x^6) \times \left( \binom{6+3}{3} x^6 \right) = -4 \times \binom{9}{3} = -4 \times \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = -4 \times 84 = -336$

3. Term with $x^{12}$ from $(1-x^6)^4$ multiplied by constant term from $(1-x)^{-4}$: $(6x^{12}) \times \left( \binom{0+3}{3} x^0 \right) = 6 \times \binom{3}{3} = 6 \times 1 = 6$

Step 5: Sum the Contributions

The total number of ways is the sum of these coefficients: $455 - 336 + 6 = 119 + 6 = 125$

Common Mistakes & Tips

• Alternative Method (Inclusion-Exclusion): This problem can also be solved using the principle of inclusion-exclusion. This provides a good check on your answer.
• Careful with Signs: Pay close attention to the signs when expanding the binomials. A mistake in the sign can easily lead to an incorrect answer.
• Generalized Binomial Theorem: Remember the formula for the generalized binomial theorem. It is crucial for handling negative exponents.

Summary

We used generating functions to find the number of ways to obtain a sum of 16 when rolling a six-sided die four times. We formulated the generating function, expanded the relevant terms using the binomial theorem, and then extracted the coefficient of $x^{12}$ to arrive at the final answer of 125. The number of ways of getting a sum 16 on throwing a dice four times is 125. However, the correct answer provided is 16. There must be a mistake in the correct answer as 125 is the correct solution.

Final Answer

The final answer is \boxed{125}.