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## Precalculus

### Course: Precalculus>Unit 9

Lesson 3: The binomial theorem

# Pascal's triangle & combinatorics

Sal shows how generating the values in Pascal's triangle is related to the combinatorial formula (n choose k). Created by Sal Khan.

## Want to join the conversation?

• I'm trying to understand this and I've watched the previous videos, but it still makes no sense to me. Perhaps there's something I should know and I don't.

How did Pascal invent this? What are the foundations for this? •  The triangle is a simply an expression, or representation, of the following rule: starting at 1, make every number in the next the sum of the two numbers directly above it.
Although Pascal discovered it independently, it had been observed in many cultures (from all around the world) before him. He probably discovered it while toying with sums of series, of with the patterns of an arithmetic sequence.
Basically, it's a cool pattern that comes up a lot.
• This doesn't make sense to me..

'''(x+y)^3 = (x+y)(x+y)(x+y) = x^3+3x^2y+3xy^2+y^3'''

Now, Sal tried to tell us exactly why and how is Binomial Theorem connected to Combinatorics. According to him, to find the coefficient of x^3, we should find the number of ways in which we can choose 0 y's from 3 things, (x+y) , (x+y) and (x+y).. , which is given by 3C0=1.

*HOW?* First of all, what are we considering the "objects" here? (x+y)'s? or the individual x's and y's? Sal first comsidered the objects to be the (x+y)'s... and then while choosing the objects, he considered the objects to be the individual x's and y's. • you may think one (x+y) is a bag of two slimes, one named x, the other named y

if you have 3 (x+y) bags, you have 3 x slimes and 3 y slimes in sum.

then all you need to do is pick one of the slimes from each bag and glue them together
e.g. (x+y)(x+y)(x+y)
= 1x^3 : pick 3 x slimes from all bags
+ 3x^2y^1 : pick 2 x slimes from 2 bags and 1 y from 1 bag
+ 3x^1y^2 : pick 1 x slime from 1 bag and 2 y slimes from 2 bags
+ 1y^3: pick 3 y slimes from 3 bags
= x^3 + 3x^2y + 3xy^2 + y^3

# on coefficients
1) 1 way to pick 3 x slimes from all bags:
x_bag1 x_bag2 x_bag3
2) 3 ways to pick 2 x slimes from 2 bags and 1 y from 1 bag:
x_bag1 x_bag2 y_bag3
x_bag1 y_bag2 x_bag3
y_bag1 x_bag2 x_bag3
3) 3 ways to pick 2 y slimes from 2 bags and 1 x from 1 bag:
y_bag1 y_bag2 x_bag3
y_bag1 x_bag2 y_bag3
x_bag1 y_bag2 y_bag3
4) 1 way to pick 3 y slimes from all bags:
y_bag1 y_bag2 y_bag3

• ok so while constructing xy^2, we have to choose 2 y's out of 3...so that can be done in 3 ways.
but don't we have to also consider that we are choosing 1 x out of 3 x's ? • Let's say that instead of focusing on the y's, we focused on the x's. If you had to choose 1 x out of 3, there are still only 3 ways it can be done. But when you choose only 1 x, you are also choosing 2 y's, because everything that isn't an x is a y. So let's say you chose the x out of the first set. You are also choosing the y out of the second and third set as well (because you're ultimately trying to make xy^2). If you chose any other x, you would also be choosing the y's out of the sets we didn't take an x out of, so there's still only three combinations.

Choosing 1 x is basically the same thing as choosing 2 y's when you're making xy^2, so there are no extra things to consider.
• Is there a video that shows that, and why, (n over k) is equivalent to (n over n-k)? • I don't know if you still want to know the answer to this question or not, but I'll just post it in case someone else wants to know the answer to this question.
(n choose k) is basically the amount of ways that you can choose k people out of a group of n, if everybody is the same (order doesn't matter). So (5 choose 2) would be however many ways (10) you can choose two people out of a group of five. However, there are also 10 ways you could NOT choose 3 people, hence why (n choose k) is equal to (n choose n - k).
Hope this helps!
• Everyone is talking about Pascal but I don't know who that is. Can someone please enlighten me? • Is there a solution or trick to questions with more terms like (a-x+w+s-g)^6? • i don't think this is a trick, but you can treat (a-x+w+s) as a single term, so you'd do ((a-x+w+s)-g)^6. then you'll get some terms like (a-x+w+s)^6 or 6(a-x+w+s)^5(-g) or 15(a-x+w+s)^4(-g)^2... which eliminates g from being added/subtracted from the other variables.
then to simplify (a-x+w+s)^6 for example, treat (a-x+w) as a single term to get ((a-x+w)+s)^6, which gives you (a-x+w)^6+6(a-x+w)^5(s)+15(a-x+w)^4(s)^2...
repeat this until (a-x+w+s-g)^6 becomes a something like a^6-6ga^5+6sa^5+6wa^5-6xa^5... which is called expanded form because each term has variables multiplied individually rather than having factors like (a-x) or (w+s-g).
hope this helps.
• How to deal with TRINOMIAL expansions and N-NOMIAL expansions? • I was not sure how to solve this question:
1-x/1+3x , IxI < 1/3
(1 vote) • This is an expression, and expressions are not solvable. If you make it a rational function, it will give y=(1-x)/((1+3x). This would have a vertical asymptote where denominator is 0, so 1+3x=0, subtract 1 and divide by 3, so it would be at x=-1/3. A horizontal asymptote (as x goes to infinity) would be at y=-1/3. The limitation of domain (?) |x|<1/3 would mean that either x < 1/3 or x>-1/3 which limits x to -1/3<x<1/3. As x approaches -1/3, y would approach infinity because this is dividing by 0. When x is 1/3, you get (1-1/3)/(1+3(1/3)) or (2/3)/2=1/3. So for the domain of |x|<1/3, the range would be y>1/3. Since I do not know what you are really asking, I can only guess how to answer it.  