(a) We can count the number of ordered pairs $(a,b)$ in two ways. First, we can simply note that there are $n$ choices for $a$ and $n$ choices for $b,$ giving a total of $n^2$ ordered pairs.
Alternatively, we can count the number of ordered pairs $(a,b)$ by dividing into cases based on the value of $a.$ For each $a,$ there are $n$ choices for $b$ (namely, $b$ can be any integer between 1 and $n,$ inclusive). Thus, the total number of ordered pairs is $\sum_{a=1}^n n = n^2.$
On the other hand, we can also count the number of ordered pairs $(a,b)$ by first choosing two distinct integers from the set $\{1,2,\ldots,n\}$ and then ordering them. There are $\binom{n}{2}$ ways to choose two distinct integers from the set, and once we have chosen them, there are two ordered pairs corresponding to them (namely, $(a,b)$ and $(b,a)$). Thus, the total number of ordered pairs is $2\binom{n}{2}.$
Since we have counted the same quantity in two different ways, we must have
\[n^2 = 2\binom{n}{2} + n.\]
(b) We can count the number of ordered triples $(a,b,c)$ in three ways.
First, we can simply note that there are $n$ choices for each of $a,$ $b,$ and $c,$ giving a total of $n^3$ ordered triples.
Alternatively, we can count the number of ordered triples $(a,b,c)$ by dividing into cases based on the values of $a$ and $b.$ For any given pair $(a,b),$ there are $n$ choices for $c,$ so the total number of ordered triples is $\sum_{a=1}^n \sum_{b=1}^n n = n^3.$
On the other hand, we can also count the number of ordered triples $(a,b,c)$ by first choosing three distinct integers from the set $\{1,2,\ldots,n\}$ and then ordering them. There are $\binom{n}{3}$ ways to choose three distinct integers from the set, and once we have chosen them, there are $3! = 6$ ordered triples corresponding to them (namely, $(a,b,c),$ $(a,c,b),$ $(b,a,c),$ $(b,c,a),$ $(c,a,b),$ and $(c,b,a)$). Thus, the total number of ordered triples is $6\binom{n}{3}.$
Finally, we can count the number of ordered triples $(a,b,c)$ by dividing into cases based on how many of $a,$ $b,$ and $c$are equal. If all three are equal, there are $n$ choices for each of $a,$ $b,$ and $c,$ giving a total of $n$ ordered triples. If exactly two are equal, there are $3n(n-1)$ choices for $(a,b,c)$ (namely, we can choose the two equal values in $n$ ways, and then choose the distinct value in $n-1$ ways). If all three are distinct, there are $6\binom{n}{3}$ choices for $(a,b,c)$ (as before). Thus, the total number of ordered triples is
\[n + 3n(n-1) + 6\binom{n}{3}.\]
Since we have counted the same quantity in three different ways, we must have
\[n^3 = n + 3n(n-1) + 6\binom{n}{3}.\]