$5$ kids toss one die each. Find the number of ways that the sum of dice points is $22$.

Question

I have already done these following steps and I'm already lost. Can someone please help me?

$x_1 + x_2 + x_3 + x_4 + x_5 = 22$ when $1 \leq x_n \leq 6$ and $n = 1,2,3,4,5$. Then, \begin{align*} f(x) & = (x+x^2+x^3+x^4+x^5+x^6)^5\\ & = \left(\frac{x(1-x^6)}{1-x}\right)^5\\ & = x^5 \cdot (1-x^6)^5 \cdot (1-x)^{-5} \end{align*}

Answer 1

By using mathematica, we have (already correct): $$x^5(1+x+x^2+x^3+x^4+x^5)^5$$ $$=x^5(x^{25}+5 x^{24}+15 x^{23}+35 x^{22}+70 x^{21}+126 x^{20}+205 x^{19}+305 x^{18}+420 x^{17}+540 x^{16}+651 x^{15}+735 x^{14}+780 x^{13}+780 x^{12}+735 x^{11}+651 x^{10}+540 x^9+420 x^8+305 x^7+205 x^6+126 x^5+70 x^4+35 x^3+15 x^2+5 x+1)$$ $$ = ... + 420x^{22}+ ...$$

so the answer is: 420

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Because: $$f(x) = x^5+(1+x+x^2+x^3+x^4+x^5)^5$$ $$=x^5[(1+x)(1+x^2+x^4)]^2$$ $$=x^5(1+x)^5[1+x^2(x^2+1)]^5$$

Note that degree of $x^k$ in $[1+x^2(x^2+1)]^5$ is even, odd degree term should be choosing from $(1+x)^5$, that is $x,x^3,x^5$, and corresponding term from $[1+x^2(x^2+1)]^5 $ is $x^{16},x^{14},x^{12}$

therefor: $$[1+x^2(1+x^2)]^5 = 1+C_5^1x^2(1+x^2)+C_5^2x^4(1+x^2)^2+C_5^3x^6(1+x^2)^3+C_5^4x^8(1+x^2)^4+C_5^5x^{10}(1+x^2)^5$$ $$=......+ C_5^3C_3^1x^{12} + (C_5^4C_4^3+C_5^5C_5^2)x^{14}+(C_5^4C_4^4+C_5^5C_5^3)x^{16}+......$$

So, the answer is:

$$C_5^5 \cdot C_5^3C_3^1 + C_5^3 \cdot (C_5^4C_4^3+C_5^5C_5^2) + C_5^1 \cdot (C_5^4C_4^4+C_5^5C_5^3)$$ $$ = 420$$

Answer 2

Method 1: You correctly found that the generating function is $$f(x) = \frac{x^5(1 - x^6)^5}{(1 - x)^5}$$ We must find the coefficient of $x^{22}$. By the Binomial Theorem, $$(1 - x^6)^5 = 1 - 5x^6 + 10x^{12} - 10x^{18} + 5x^{24} - x^{30}$$ Hence, \begin{align*} f(x) & = \frac{x^5(1 - 5x^6 + 10x^{12} - 10x^{18} + 5x^{24} - x^{30}}{(1 - x)^5}\\ & = \frac{x^5 - 5x^{11} + 10x^{17} - 10x^{23} + 5x^{29} - x^{35}}{(1 - x)^5} \end{align*} If $n$ is a positive integer, then the coefficient of $x^k$ in the expansion of $(1 - x)^{-n}$ is $$\binom{n + k - 1}{k} = \binom{n + k - 1}{n - 1}$$ Hence, $$\frac{1}{(1 - x)^5} = \sum_{k = 0}^{n} \binom{k + 4}{4} x^k$$ Therefore, $$f(x) = (x^5 - 5x^{11} + 10x^{17} - 10x^{23} + 5x^{29} - x^{35})\sum_{k = 0}^{\infty} \binom{k + 4}{4} x^k$$ Notice that only the $k = 17, 11, 5$ terms contribute to the $x^{22}$ coefficient. Therefore, the number of ways to obtain a sum of $22$ when five dice are thrown is $$\binom{17 + 4}{4} - 5\binom{11 + 4}{4} + 10\binom{5 + 4}{4} = 420$$

Method 2: We can confirm this result by using the Inclusion-Exclusion Principle.

The number of solutions of the equation $$x_1 + x_2 + x_3 + \ldots + x_k = n$$ in the positive integers is $$\binom{n - 1}{k - 1}$$ Hence, the number of solutions of the equation $$x_1 + x_2 + x_3 + x_4 + x_5 = 22 \tag{1}$$ in the positive integers is $$\binom{22 - 1}{5 - 1} = \binom{21}{4}4$$ From these, we must exclude those solutions in which one or more of the variables exceeds $6$. Notice that since $3 \cdot 7 + 2 = 23 > 22$, at most two of the variables may exceed $6$.

Suppose $x_1 > 6$. Then $x_1' = x_1 - 6$ is a positive integer. Substituting $x_1' + 6$ for $x_1$ in equation 1 and simplifying yields \begin{align*} x_1' + 6 + x_2 + x_3 + x_4 + x_5 + x_6 & = 22\\ x_1' + x_2 + x_3 + x_4 + x_5 + x_6 & = 16 \tag{2} \end{align*} Equation 2 is an equation in the positive integers with $$\binom{16 - 1}{5 - 1} = \binom{15}{4}$$ solutions. By symmetry, there are an equal number of solutions for each variable that could exceed $6$. Hence, there are $$\binom{5}{1}\binom{15}{4}$$ solutions in which one of the variables exceeds $6$.

However, we have subtracted too much since we have subtracted those cases in which two of the variables exceed six twice, once for each of the ways we could have designated a particular variable as the one that exceeds six. Since we only want to exclude these cases once, we must add them back.

Suppose $x_1, x_2 > 6$. Then $x_1' = x_1 - 6$ and $x_2' = x_2 - 6$ are positive integers. Substituting $x_1' + 6$ for $x_1$ and $x_2' + 6$ for $x_2$ in equation 1 yields \begin{align*} x_1' + 6 + x_2' + 6 + x_3 + x_4 + x_5 & = 22\\ x_1' + x_2' + x_3 + x_4 + x_5 & = 10 \tag{3} \end{align*} Equation 3 is an equation in the positive integers with $$\binom{10 - 1}{5 - 1} = \binom{9}{4}$$ solutions. By symmetry, there are an equal number of solutions for each of the $\binom{5}{2}$ ways we could select two variables to exceed six. Hence, there are $$\binom{5}{2}\binom{9}{4}$$ solutions in which two of the variables exceed six.

By the Inclusion-Exclusion Principle, the number of ways the sum of the points on the five dice could equal $22$ is $$\binom{21}{4} - \binom{5}{1}\binom{15}{4} + \binom{5}{2}\binom{9}{4}$$

Answer 3

As with the other answers, here is an expansion of the generating function. $$ \begin{align} x^5\left(\frac{1-x^6}{1-x}\right)^5 &=x^5\sum_{j=0}^5\binom{5}{j}\left(-x^6\right)^j\sum_{k=0}^\infty\binom{-5}{k}(-x)^k\tag{1}\\ &=\sum_{j=0}^5\sum_{k=0}^\infty(-1)^{j+k}\binom{5}{j}\binom{-5}{k}x^{k+6j+5}\tag{2}\\ &=\sum_{j=0}^5\sum_{k=0}^\infty(-1)^j\binom{5}{j}\binom{k+4}{k}x^{k+6j+5}\tag{3}\\ &=\sum_{j=0}^5\sum_{k=6j+5}^\infty(-1)^j\binom{5}{j}\binom{k-6j-1}{k-6j-5}x^k\tag{4}\\ &=\sum_{k=5}^\infty\color{#C00}{\sum_{j=0}^{\left\lfloor\frac{k-5}6\right\rfloor} (-1)^j\binom{5}{j}\binom{k-6j-1}{4}}x^k\tag{5} \end{align} $$ Explanation:
$(1)$: Binomial Theorem
$(2)$: combine and rearrange
$(3)$: $\binom{-5}{k}=(-1)^k\binom{k+4}{k}$
$(4)$: $k\mapsto k-6j-5$
$(5)$: change order of summation and $\binom{n}{k}=\binom{n}{n-k}$

For $k=22$, we get from the red part of $(5)$: $$ \begin{align} \sum_{j=0}^2 (-1)^j\binom{5}{j}\binom{21-6j}{4} &=\binom{5}{0}\binom{21}{4}-\binom{5}{1}\binom{15}{4}+\binom{5}{2}\binom{9}{4}\\[6pt] &=420 \end{align} $$