Time complexity of a triple-nested loop

Please consider the following triple-nested loop:

for (int i = 1; i <= n; ++i)
    for (int j = i; j <= n; ++j)
        for (int k = j; k <= n; ++k)
            // statement

The statement here is executed exactly $n(n+1)(n+2)\over6$ times. Could someone please explain how this formula was obtained? Thank you.

You can count the number of times the innermost for loop is executed by counting
the number of triplets $(i,j,k)$ for which it is executed.

By the loop conditions we know that: $1 \leq i \leq j \leq k \leq n$ . We can reduce it to the following simple combinatorics problem.

• Imagine $n+2$ boxes of red colour placed in an array from left to right.

• Pick any 3 boxes from the $n+2$ boxes and paint them blue.

• Form a triplet $(i,j,k)$ as follows:

• $i$ = 1 + number of red coloured boxes to the left of first blue box.

• $j$ = 1 + number of red coloured boxes to the left of second blue box.

• $k$ = 1 + number of red coloured boxes to the left of third blue box.

So, we just need to count the number of ways of picking 3 boxes from $n+2$ boxes which is $n+2 \choose 3$.

StackExchange Computer Science Q#3306, answer score: 14