Pascal's triangle I was out sick for a while (2 weeks) and just got back and now we are doing whatever this is! Can someone explain to me what this is or show me a video on how to do it?
"Use Pascal's triangle and the regularity of decreasing powers of $a$ and $b$ of increasing powers to develop the expression $(4a-3b) ^ 6$"
 A: Pascal's triangle is based on an identity for choosing unordered samples taken without replacement (no item sampled more than once).
Binomial coefficients. The number of unordered samples of size $k$ from a larger set of
size $n$ is $$C(n, k) = {}_n C_k = {n 
\choose k} = \frac{n!}{k!(n-k)!},$$
for integers $0 \le k \le n.$ I have shown three common notations for this, and will use $C(n, k)$ below. Sometimes numbers $C(n,k)$ are called binomial coefficients.
For example, suppose $n = 4$ and $k = 2$. Let the four objects
be denoted A, B, C, and D. Then there are $4!/[2!(4-2)!] = 24/4 = 6$ ways to choose two of the objects, without regard to order. Because actual order is not important, I will list them in alphabetical order: AB, AC, AD, BC, BD, CD. Sometimes, these numbers can be quite large. For example the number of possible 5-card poker hands chosen from a 52-card deck is $C(52, 5) = 2,598,960,$ which I computed using software.
Pascal's Triangle. The equation on which Pascal's Triangle is
based is the following;
$$C(n, k) = C(n-1, k-1) + C(n-1, k).$$
An example, is $C(5, 3) = C(4, 2) + C(4, 3),$ which amounts to
$10 = 6 + 4$ after you do the computation of each binomial coefficient.
One can give a mathematical proof of Pascal's Triangle. However, I will give a simple sampling argument for it. Suppose I am teaching
a class of 30 students, and 4 out of the 31 of us (30 students plus me) will be given \$20 gift certificates. There are $C(31, 4)$ ways to do this. But from my (selfish) point of view, there are two kinds of outcomes, ones in which I get one of the prizes and ones in which I don't. There are $C(30, 3)$ ways to make the choices so I get a prize: give me my prize then figure out how many ways to give 3 prizes to the thirty students. There are $C(30, 4)$ ways to do the choices so I don't get a prize: send me out of the room and pick 4 students to get the prizes. Obviously, we must have
$$C(31, 4) = C(30, 3) + C(30, 4).$$
Upon computation this becomes $31,465 = 4060 + 27,405.$
I will let you look at one of the sources available online to see how binomial coefficients are arranged into a large triangle. The idea is that the numbers on the $n$th row of the triangle are found by summing two adjacent numbers in row $n-1.$
Each row begins and ends with a $1$ because $C(n, 0) = C(n, n) = 1.$
The second element in a row tells you the row number because $C(n,1) = n.$ Each row is one element longer than the one above it. In theory the process can go on forever; there is never a last row. Here are the first few:
                    1
                 1     1
              1     2     1
          1      3     3     1
       1      4     6     4     1
    ................................

Can you find where $C(3, 1) + C(3, 2) = C(4, 2)$ fits into the last two rows above? 
Your question about a binomial expansion uses binomial coefficients
from the 6th row (not shown):
$$(4a - 3b)^6 = C(6,0)(4a)^6(-3b)^0 + C(6,1)(4a)^5(-3b)^1 + \cdots
+ C(6,6)(4a)^0(-3b)^6.$$
BT
