Find the number of tuples consisting of $0, 1$ and $3$

We can successively use the binomial theorem to extract the coefficient. It is convenient to use the coefficient of operator $[x^k]$ to denote the coefficient of $x^k$ in a polynomial.

We obtain \begin{align*} [x^{15}](1 + x + x^3)^{26}&=[x^{15}]\sum_{k=0}^{26}\binom{26}{k}x^{3k}(1+x)^{26-k}\\ &=\sum_{k=0}^{5}\binom{26}{k}[x^{15-3k}](1+x)^{26-k}\tag{1}\\ &=\sum_{k=0}^{5}\binom{26}{k}[x^{15-3k}]\sum_{j=0}^{26-k}\binom{26-k}{j}x^{j}\\ &=\sum_{k=0}^{5}\binom{26}{k}\binom{26-k}{15-3k}\tag{2}\\ &=853423740 \end{align*}

Comment:

• In (1) we use the linearity of the coefficient of operator and the rule $$[x^{n-k}]P(x)=[x^n]x^kP(x)$$ We also set the upper limit of the index $k$ to $5$, since the exponent of $x^{15-3k}$ is non-negative.

• In (2) we select the summand with $j=15-3k$ to obtain the coefficient of $x^{15-3k}$.

Okay, so you have options: Lets say you have $a$ 1s and $b$ 3s, and the remainder are zeros. So you have that $a+3b = 15$. This gives the following collection of $(a,b)$ pairs as solutions: $$ (15,0), (12, 1), (9,2), (6,3), (3,4), (0,5),$$ or equivalently, $(15-3k, k)$ for $0 \leq k \leq 5$.

Now from a pair $(a,b)$ you obtain solutions to the original equation by choosing, from the 26 positions, $a$ places to put a 1, and from the remaining $26-a$ positions, $b$ places to put a $0$. In other words, you have contribution of ${26 \choose a}{26-a \choose b}$ from the pair $(a,b)$. This gives a total of $$ \sum_{k=0}^{5} {26 \choose 15 -3k}{26-3k \choose k}.$$ This gives the same result as in the answer of Markus.