The standard notation for the number in row $n$ and column $k$ is $\left ( \begin{array}{c} n\\k \end{array} \right )$. So $\left ( \begin{array}{c} n\\0 \end{array} \right )$ is always $1$, and $\left ( \begin{array}{c} n\\n \end{array} \right )$ is always $1$, whereas $\left ( \begin{array}{c} 5\\3 \end{array} \right ) = 10$.
The entries in Pascal's triangle appear in the Binomial Theorem:
$$(a+b)^n = \left ( \begin{array}{c} n\\0 \end{array} \right ) a^n +
\left ( \begin{array}{c} n\\1 \end{array} \right ) a^{n-1}b +
\left ( \begin{array}{c} n\\2 \end{array} \right ) a^{n-2}b^2 + \dots +
\left ( \begin{array}{c} n\\n-2 \end{array} \right ) a^2b^{n-2}+
\left ( \begin{array}{c} n\\n-1 \end{array} \right ) ab^{n-1} +
\left ( \begin{array}{c} n\\n \end{array} \right ) b^n.$$
So for example
$$(a+b)^0 = 1$$
$$(a+b)^1 = a+b$$
$$(a+b)^2 = a^2 + 2ab + b^2$$
$$\dots$$
$$(a+b)^5 = a^2 + 5ab^4 + 10a^2b^3 + 10 a^3b^2 + 5a^4b + b^5.$$
You can check that $$(1+1)^5 = 1 + 5 + 10 + 10 + 5 + 1$$ or $2^5 = 32$. More generally, a similar calculation shows that the sum of the numbers in row $n$ is $2^n$. Make sure you understand why this follows from the Binomial Theorem.
$1$
$1$$1$
$1$$0$$1$
$1$$1$$1$$1$
$1$$0$$0$$0$$1$
$1$$1$$0$$0$$1$$1$
$1$$0$$1$$0$$1$$0$$1$
$1$$1$$1$$1$$1$$1$$1$$1$
$1$$0$$0$$0$$0$$0$$0$$0$$1$
$1$$1$$0$$0$$0$$0$$0$$0$$1$$1$
$1$$0$$1$$0$$0$$0$$0$$0$$1$$0$$1$
$1$$1$$1$$1$$0$$0$$0$$0$$1$$1$$1$$1$
$1$$0$$0$$0$$1$$0$$0$$0$$1$$0$$0$$0$$1$
$1$$1$$0$$0$$1$$1$$0$$0$$1$$1$$0$$0$$1$$1$
$1$$0$$1$$0$$1$$0$$1$$0$$1$$0$$1$$0$$1$$0$$1$
$1$$1$$1$$1$$1$$1$$1$$1$$1$$1$$1$$1$$1$$1$$1$$1$
etc.
Notice that the triangle formed by the first 16 rows (number 0 to 15) contains three smaller triangles of height 8 and of the form:
$1$
$1$$1$
$1$$0$$1$
$1$$1$$1$$1$
$1$$0$$0$$0$$1$
$1$$1$$0$$0$$1$$1$
$1$$0$$1$$0$$1$$0$$1$
$1$$1$$1$$1$$1$$1$$1$$1$
and that each of them contains three triangles of height 4 and of the form:
$1$
$1$$1$
$1$$0$$1$
$1$$1$$1$$1$
and that each of those contains three triangles of height 2 and of the form:
$1$
$1$$1$
and that each of them contains three "triangles" of height 1!
1
11
121
1001
11011
121121
1002001
11022011
121212121
1000000001
etc.
Note that the entries in this table can be calculated directly using addition modulo 3: \begin{array}{c|ccc} +&0&1&2\\\hline 0&0&1&2\\ 1 &1&2&0\\ 2&2&0&1 \end{array} Can you see a Sierpinski-like process here? Explain. Tell as much about it as you can. Feel free to calculate additional rows!
Check that this definition also gives $\left ( \begin{array}{c} 5\\3 \end{array} \right ) = 10$.
Show that the coefficients defined using this definition satisfy: $$\left ( \begin{array}{c} n+1\\k+1 \end{array} \right ) = \left ( \begin{array}{c} n\\k \end{array} \right ) + \left ( \begin{array}{c} n\\k+1 \end{array} \right ).$$ Hint: put the terms on a common denominator.
Remember: Collaboration is fine, but what you hand in must be in your own words. Handing in something you copied is plagiarism and will cost you if it is detected. Write down what you tried and how it worked.
Hand in your work at the recitation meeting (Mon, Wed or Thur)
next week.