Binomial theorem yo

Generally speaking, $\sum_{K \leq N} \binom N K a^K = \sum_{K\leq N} \binom N K a^K 1^{N-K} = (1 + a)^N.$ Apply this three times: $\sum_{i \leq j} \binom j i = (1 + 1)^j = 2^j;$ $\sum_{j \leq n} \sum_{i \leq j} \binom n j \binom j i = \sum_{j \leq n} \binom n j 2^j = (2 + 1)^n = 3^n;$ $\sum_{n \leq 10} \sum_{j \leq n} \sum_{i \leq j} \binom {10} n \binom n j \binom j i = \sum_{n \leq 10} \binom {10} n 3^n = (3 + 1)^{10} = 4^{10}.$ Therefore $K = 4^{10} = 2^{20}$ . Its 21 positive divisors are $2^k$ with $k = 0, \dots, 20$ ; only $\boxed{20}$ of these are proper divisors.

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A combinatorial approach: We can interpret the sum as describing the following process:

Out of 10 people, we choose $n$ to be interviewed. Out of these $n$ , we choose $j$ to be hired. Out of these $j$ , we choose $i$ to receive tenure. The sum counts all possible ways of doing this.

Now each of the 10 people is one of four things: not interviewed; interviewed but not hired; hired but not tenured; or tenured. This makes for $4^{10}$ possibilities.