<s>[INST] Sum of the rows of Pascal's Triangle. I've discovered that the sum of each row in Pascal's triangle is $2^n$, where $n$ number of rows. I'm interested why this is so. Rewriting the triangle in terms of C would give us 
              $0C0$ in first row.
$1C0$ and $1C1$ in the second, and so on and so forth. However, I still cannot grasp why summing, say, 4C0+4C1+4C2+4c3+4C4=2^4. 
 [/INST] There are various different ways to look at this.  Here's one:
Two adjacent numbers in a row get added to get the number in the row below it:
$$
\begin{array}{cccccccccc}
& & 1 & & & & & 8 & & & & 28 & & & & 56 & & & & 70 & & \cdots \\
& & & & & & & & \searrow & & \swarrow  \\
1 & & & & & 9 & & & & 36 & & & & 84 & & & & 126 & & \cdots & & & & \cdots
\end{array}
$$
That means every number in a row is added into the next row twice:
$$
\begin{array}{cccccccccc}
& & 1 & & & & & 8 & & & & 28 & & & & 56 & & & & 70 & & \cdots \\
& & & & & & \swarrow & &  \searrow  \\
1 & & & & & 9 & & & & 36 & & & & 84 & & & & 126 & & \cdots & & & & \cdots
\end{array}
$$
Since every number is added into the next row twice, the sum of the numbers in the next row is twice as big.
Here's another:  In row $9$ (which is the tenth row, since the first row is "row $0$), the entries are.
$$
\binom 9 0 = 1,\  \binom 9 1 = 9,\  \binom 9 2 = 36,\  \binom 9 3 = 84,\  \binom 9 4 = 126,\ \ldots
$$
These are


*

*the number of subsets of size $0$ of a set of size $9$, and

*the number of subsets of size $1$ of a set of size $9$, and

*the number of subsets of size $2$ of a set of size $9$, and

*the number of subsets of size $3$ of a set of size $9$, and

*and so on.


Their sum is therefore the total number of subsets of a set of size $9$.  If you know that that is $2^9$, you've got it.
</s>