Binomial coefficients galore

We interpret each sum $\sum_{k=0}^n \binom n k$ as choosing an arbitrary number $k$ out of $n$ items of a set, and counting in how many ways that may be done. Extending this idea gives an interpretation for the value of $N$ : Consider a collection of 50 sheets of paper, which are initially blank.

First, choose $a$ out of these 50 sheets and draw a circle on each of them.

Of the $a$ sheets with a circle, pick $b$ and draw a square on each of them.

Of the $a$ sheets with a circle, pick $c$ and draw a triangle on each of them.

Of the $50 - a$ sheets without a circle, pick $d$ and draw an semicircle on each of them.

Of the $c$ sheets with a triangle and the $d$ sheets with a semicircle, pick $e$ and draw a star on each of them.

Of the $50 - a - d$ sheets without circle or semicircle, pick $f$ sheets and draw a dot on them.

After this process, each of the 50 sheets belongs to precisely one of the following categories:

• only a circle

• a circle and a square

• a circle and a triangle

• a circle, a triangle, and a star

• a circle, a square, and a triangle

• a circle, a square, a triangle, and a star

• only a semicircle

• a semicircle and a star

• only a dot

• blank

Since each of the 50 sheets belongs to precisely one out of 10 categories, there are $N = 10^{50}$ ways of making these choices. This number has $\boxed{51}$ digits.

For any $a,d \ge 0$ we have $\sum_{b=0}^a \sum_{c=a}^a \sum_{e=0}^{c+d} \binom{a}{b}\binom{a}{c}\binom{c+d}{e} \; = \; \sum_{b=0}^a \sum_{c=0}^a\binom{a}{b}\binom{a}{c}2^{c+d} \; = \; 2^a \times 3^a \times 2^d \; = \; 6^a \times 2^d$ and hence $N \; = \; \sum_{a+d+f \le 50} \frac{50!}{a!\, d! \, f! \, (50-a-d-f)!} \times 6^a \times 2^d \; = \; (6 + 2 + 1 + 1)^{50} \; = \; 10^{50}$ which has $\boxed{51}$ digits.

By repeated use of the Binomial Theorem, $\begin{aligned} &\sum_{a = 0}^{50} \sum_{b = 0}^a \sum_{c = 0}^a \sum_{d = 0}^{50 - a} \sum_{e = 0}^{c + d} \sum_{f = 0}^{50 - a - d} \binom{50}{a} \binom{a}{b} \binom{a}{c} \binom{50 - a}{d} \binom{c + d}{e} \binom{50 - a - d}{f} \\ &= \sum_{a = 0}^{50} \sum_{b = 0}^a \sum_{c = 0}^a \sum_{d = 0}^{50 - a} \sum_{e = 0}^{c + d} \binom{50}{a} \binom{a}{b} \binom{a}{c} \binom{50 - a}{d} \binom{c + d}{e} 2^{50 - a - d} \\ &= \sum_{a = 0}^{50} \sum_{b = 0}^a \sum_{c = 0}^a \sum_{d = 0}^{50 - a} \binom{50}{a} \binom{a}{b} \binom{a}{c} \binom{50 - a}{d} 2^{c + d} \cdot 2^{50 - a - d} = \sum_{a = 0}^{50} \sum_{b = 0}^a \sum_{c = 0}^a \sum_{d = 0}^{50 - a} \binom{50}{a} \binom{a}{b} \binom{a}{c} \binom{50 - a}{d} 2^{50 + c - a} \\ &= \sum_{a = 0}^{50} \sum_{b = 0}^a \sum_{c = 0}^a \binom{50}{a} \binom{a}{b} \binom{a}{c} 2^{50 - a} \cdot 2^{50 + c - a} = \sum_{a = 0}^{50} \sum_{b = 0}^a \sum_{c = 0}^a \binom{50}{a} \binom{a}{b} \binom{a}{c} 2^{100 + c - 2a} \\ &= \sum_{a = 0}^{50} \sum_{b = 0}^a \binom{50}{a} \binom{a}{b} 2^{100 - 2a} \sum_{c = 0}^a \binom{a}{c} 2^c = \sum_{a = 0}^{50} \sum_{b = 0}^a \binom{50}{a} \binom{a}{b} 2^{100 - 2a} \cdot 3^a \\ &= \sum_{a = 0}^{50} \binom{50}{a} 2^a \cdot 2^{100 - 2a} \cdot 3^a = \sum_{a = 0}^{50} 2^{100 - a} \cdot 3^a \\ &= 2^{50} \sum_{a = 0}^{50} 2^{50 - a} \cdot 3^a = 2^{50} \cdot 5^{50} = 10^{50}, \end{aligned}$ which has 51 digits.