Problem: Compute $ \sum_{n_{60}=0}^2 \sum_{n_{59}=0}^{n_{60}} \cdots \sum_{n_2=0}^{n_3} \sum_{n_1=0}^{n_2} \sum_{n_0=0}^{n_1} 1 . $ Solution: If you think you are going to solve this algebraically, good luck. This problem is screaming at you to double count. Consider the set of length-61 ternary strings that are decreasing. If we pick a distribution of 61 $0s, $1s, and $2s, then the ordered string is determined, so the number of such strings is $\binom{61 + 3 - 1}{3-1} = \binom{63}{2}$. Now, consider the following alternate way of counting this set of ternary strings. Look at the first digit. The value we choose (between $0,1,2$) for this first digit opens up that exact amount of possibilities. Say we choose $0$. Then, we have exactly $0$ ways to assign the next digit. For each of those ways, we repeat. In another case, if the first digit is $2$, then for the next digit, it can either be $0,1,2$, **for each** value it is, we assign it and then carry on. This "for each" language directly corresponds to the sum above. If that is not rigorous enough, consider the fact that the sum above can also be written $\sum_{n_{60}=0}^2 \sum_{n_{59}=0}^{n_{60}} \cdots \sum_{n_2=0}^{n_3} \sum_{n_1=0}^{n_2} \sum_{n_0=0}^{n_1} 1 = \sum_{2 \geq n_{60} \geq n_{59} \geq \dots \geq n_{1} \geq n_{0} \geq 0} 1$which now directly corresponds to the length-61 ternary strings whose digits are decreasing.