Let $S_n$ be the set of all (finite) sequences of length $n$, with all entries being whole numbers between $1$ and $6$. Similarly, let $B_n$ be the set of all (finite) sequences of length $n$, with all entries being either $0$ or $1$. We denote elements of $S_n$ or $B_n$ by a letter such as $x$, and a subscripted $x_i$ denotes the $i$th entry in $x$.
Let $B$ be the subset of $B_5$ that have exactly five entries equal to $1$, so the cardinality of $B$ is $\binom{15}{5}$. We now set up a bijection from $B$ to the set $S$ of nondecreasing sequences in $S_{10}$. We first set up a map $f$ from $B$ to $S_{10}$: $y = f(x)$ is such that $y_1 = 1$ and $y_{i+1} - y_i = x_i$ for $1 \le i \le 15$. This defines $f$ uniquely. $f$ is clearly injective since $y$ preserves the information about which entries in $x$ equal $1$. It is also clear that $y$ is nondecreasing.
Now, delete $y_1$ and also delete $y_i$ if $x_i = 1$. We are left with a nondecreasing sequence $z = g(y)$ of length $10$, and so $g$ is a map from $f(B)$ to $S_{10}$. A little thought shows that we can reconstruct $y$ from $z$, and so $g$ is injective. We deduce that $h := g \circ f$ is injective. $z$ is nondecreasing because $y$ is nondecreasing, and it is now straightforward to deduce that $h(B) = S$. We have constructed a bijection $h$ from $B$ to $S$, and so the cardinality of $S$ is $\binom{15}{10}$.