The problem can be solved easily using generating functions . The generating function associated with the term $3x$ is $1+x^3+x^6+...+x^{24}= \frac{1}{1-x^3}$ , while $y,z$ have correspond to the same polynomial $\frac{1}{1-x}$ . Solving the question now comes down to finding the coefficient of $x^{24}$ in the expansion of $\frac{1}{1-x^3} \cdot (\frac{1}{1-x})^2$ , which can be done using (for example) Wolfram Alpha leading to $117$ .

Doing the calculation by hand is not difficult:

$\frac{1}{(1-x)^2} = \sum_{k=0}^{\infty} {2+k-1 \choose 1} \cdot x^k = 1+2x+3x^2+4x^3+\dots+25x^{24}$

from the negative binomial series . We have to evaluate the coefficient of $x^{24}$ in the product (ignoring higher order terms) $(1+x^3+x^6+\dots +x^{24}) \cdot (1+2x+3x^2+\dots+25x^{24})$ which is equivalent to evaluating the sum $25+22+19+16+\dots+1= 9+3\cdot(1+2+\cdot+8)=9+3\cdot \frac{8\cdot9}{2}=117$

Since $x$ must be an integer, $3x$ must also be an integer. Between $0$ and $24$ these multiples $k$ are $0, 3, 6, 9, 12, 15, 18, 21, 24$ . If anyone knows a more efficient way, please let me know. I, being the lazy person, simply counted the number of solutions for each $y+z=24-k$ using stars and bars . Adding them together gives us $117$ .

The general formula for the number of solutions to an equation of the kind $rx_1+x_2+...+x_n=kr$ is $\displaystyle\sum_{i=0}^{k} \binom{r(k-i)+n-2}{n-2}$ . Can anyone give a proof?

@Brian Charlesworth How could this be done quicker?

Since I'm currently studying Python, here's my Python code:

            n=0

            for x in range(9):

            for y in range(25):

            for z in range(25):

            if x*3+y+z == 24:

            n += 1

        print n

$\color{#3D99F6}{117}$

Since $3x$ is the variable with the greatest coefficient, let us fix its value and find out the possibilities of the remaining 2. $x$ can take a value from $(0,1,2...8)$ which yields $3x$ as $(0,3,6...24)$ .

For each value of $3x$ , there exists $24-3x +1$ solutions to the ordered pair $(y,z)$ . How? Simple.

Since $y$ and $z$ both have coefficient one, the possibility of the ordered pairs would be a symmetric set with values from $0$ to $24-3x$ . Which means, there would be $24-3x + 1$ ordered pairs in each case.

Now instead of sitting and calculating each value of $24-3x+1$ , we could make a peculiar observation regarding the value of the above expression.

Listing out all possibilities of $3x$ : $0,3,6,9,12,15,18,21,24$

If you notice, this is a mirrored set of values, i.e. the $n^{th}$ value of the set corresponds to the value of $24- 3n$ which comes at the place $10 - n$ . In simpler terms, $(24-0=24)$ , $(24-3=21)$ ... $(24-24 = 0)$ . Therefore the total would be the sum of all possible values of $3x+9$ (since there is a $+1$ in each case ).

$0 + 3 + 6 + 9 + 12 + 15 + 18 + 21 + 24 + 9$ $\boxed{117}$

here the possible value of x is 0 to 8. if x=0 then y+z=24 and we have 25C1 or 25 solution. then if x=1 we have 22c1=22 solution. if this process is going on then we can reach to x=8 and y+z=0 and we have only one solution in this case. so the number of possible solution is 25+22+19+16.......+1 = 117

In the Taylor series expansion of 1/{(1-x^3)(1-x)^2} at x=0, find the coefficient of x^24, using WolframAlpha and get the Answer=117.