Find the coefficient of $x^{12}y^{24}$ in $(x^3 + 2xy^2 +y + 3)^{18}$.
I have been working on this problem for a while now and I cannot figure out how to use the multinomial theorem to solve it. I tried using auxiliary variables but then I did not know what the total exponent should be. I am really stuck on this.
On
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Find the coefficient of $\ds{x^{12}y^{24}\ \mbox{in}\ \pars{x^{3} + 2xy^{2} + y + 3}^{18}:\ {\large ?}}$
\begin{align}&\dsc{\pars{x^{3} + 2xy^{2} + y + 3}^{18}} =\sum_{\atop{a, b, c, d\ \geq\ 0 \atop\vphantom{\LARGE A} a + b + c + d\ =\ 18}}{18! \over a!\, b!\, c!\, d!} \pars{x^{3}}^{a}\pars{2xy^{2}}^{b}y^{c}3^{d} \\[5mm]&=18!\sum_{\atop{a, b, c, d\ \geq\ 0 \atop\vphantom{\LARGE A} a + b + c + d\ =\ 18}}{2^{b}3^{d} \over a!\, b!\, c!\, d!}\,x^{3a + b}y^{2b + c}\,,\qquad a, b, c, d \in {\mathbb N} \end{align}
Now, we have three equations which let us to express $\ds{a,b\ \mbox{and}\ c}$ in terms of $\ds{d}$: $$ \left.\begin{array}{rcrcrcl} a & + & b & + & c & = & 18 - d \\ 3a & + & b &&& = & 12 \\ && 2b & + & c& = & 24 \end{array}\right\}\ \imp\ \left\{\begin{array}{lcl} a & = & {1 \over 4}\,\pars{6 - d} \\[2mm] b & = & {3 \over 4}\,\pars{10 + d} \\[2mm] c & = & {3 \over 2}\,\pars{6 - d} \end{array}\right. $$ From those identities we conclude that $\ds{d = 2}$ or $\ds{d = 6}$.Namely, $$ \pars{a = 1\,,\quad b = 9\,,\quad c = 6\,,\quad d = \dsc{\large 2}}\ \mbox{and}\ \pars{a = 0\,,\quad b = 12\,,\quad c = 0\,,\quad d = \dsc{\large 6}} $$
The coefficient of $\ds{x^{12}y^{24}\ \mbox{in}\ \pars{x^{3} + 2xy^{2} + y + 3}^{18}}$ becomes: $$ 18!\pars{{2^{9}\ 3^2 \over 1!\, 9!\, 6!\, 2!} +{2^{12}\ 3^6 \over 0!\, 12!\, 0!\, 6!}} =\color{#66f}{\large 111\ 890\ 128\ 896} $$
Break it into cases based on how the $x^{12}$ was formed from some combination of $x^3$ and $2xy^2$. In each case, you must also check that it is possible that $y$ could be raised to the 24th power using the remaining available options.
Case 1: $(x^3)^4$: You might have had an $x^{12}y^{24}$ term because of $(x^3)^4\cdot (2xy^2)^0\cdot(y)^{24}\cdot 3^n$, but we're only raising to the 18th power, so that can't happen.
Case 2: $(x^3)^3$: It could have been $(x^3)^3\cdot(2xy^2)^3\cdot (y)^{18}\cdot 3^n$. Also not possible, not large enough power available.
Case 3: $(x^3)^2$: In this case it might have been $(x^3)^2\cdot (2xy^2)^6\cdot y^{12}\cdot3^n$, again not possible. It would again require too large of a power of the original function (at least being raised to 20th power)
Case 4: $(x^3)^1$: This would imply it would be $(x^3)^1\cdot (2xy^2)^9\cdot (y)^6\cdot(3)^2$ This is infact possible.
Case 5: $(x^3)^0$: This would imply that it would be $(x^3)^0\cdot (2xy^2)^{12}\cdot (y)^0\cdot (3)^6$ This also is possible.
Now you know that it could have been formed one of two ways. Reimagining the problem instead as $(a+b+c+d)^{18}$, the question is then what are the coefficients of the $ab^9c^6d^2$ and $b^{12}d^6$ terms.
The coefficient of the $ab^9c^6d^2$ term is $\dfrac{18!}{1!9!6!2!}$ and the coefficient of the $b^{12}d^6$ term is $\dfrac{18!}{12!6!}$
So, we have $\dfrac{18!}{1!9!6!2!}(x^3)^1\cdot (2xy^2)^9\cdot (y)^6\cdot(3)^2 + \dfrac{18!}{12!6!}(2xy^2)^{12}\cdot (3)^6 = Cx^{12}y^{24}$
So, $C = \dfrac{18!}{9!6!2!}\cdot 2^9\cdot3^2 + \dfrac{18!}{12!6!}\cdot 2^{12}\cdot 3^6$