Learn math, science, English SAT & ACT from
highquaility study
videos by expert teachers
Thank you for watching the preview.
To unlock all 5,300 videos, start your free trial.
Binomial Theorem  Problem 1
Carl Horowitz
Carl Horowitz
University of Michigan
Runs his own tutoring company
Carl taught upperlevel math in several schools and currently runs his own tutoring company. He bets that no one can beat his love for intensive outdoor activities!
Combinations or choosing. So this is an application of factorials that come up in algebra 2 and as well that come up in precalculus. But for right now we’re going to focus on the basic premise of choosing, that’s the definition of it. And what this is saying is n choose r, and what it comes with is this weird looking formula, okay?
And factorial, so it’s the number on the outside, factorial, divided by the difference of your two numbers factorial times the other number factorial. That’s the definition, sort of confusing on the surface but we’ll look into it a little bit more and also one thing you should know is that this is sometimes written also as parenthesis n and r. So it’s just basically n choose r. It’s a different way of writing the exact same formula we have over here.
So let’s actually put this into application. So what we have here is 9 choose 3. Putting this into our equation, what we end up having is the n factorial is the 9, this becomes 9 factorial. The n minus r so the big number minus the small number becomes 9 minus 3, this becomes 9 minus 3 factorial and lastly the r factorial, the small number factorial is just 3 factorial.
Simplifying this up, we know that we have 9 factorial and then the 9 minus 3 is just 6 factorial, 3 factorial. A factorial remember is just that number multiplied by every number smaller than that, so what we have here is 9 times 8 times 7 and so on and so forth. This is 6 times 5 times 4, so the 6 in down all cancel leaving us with a 9 times 8 times 7 in the numerator, the 6 and down all cancelled out, over 3 factorial, which is just 3 times 2 times 1. Again that 1 we don’t really need because it’s just times 1 but it’s always good to throw it in there.
Now, we just have a fraction which we can cancel terms and try to make our life a little bit easy. 9 over 3 goes to 3, 8 over 2 ends up with a 4. So what we end up with is 3 times 4 times 7 and we have nothing left in the bottom. 3 times 4 is 12, 12 times 7 is, let’s make sure we get it right, 12 times 7 is 84. So using our formula and definition, we found that 9 choose 3 is 84. So it’s a scarylooking formula but as long as you know where each term goes, it’s pretty straight forward to use.
Please enter your name.
Are you sure you want to delete this comment?
Carl Horowitz
B.S. in Mathematics University of Michigan
He knows how to make difficult math concepts easy for everyone to understand. He speaks at a steady pace and his stepbystep explanations are easy to follow.
Concept (1)
Sample Problems (8)
Need help with a problem?
Watch expert teachers solve similar problems.

Binomial Theorem
Problem 1 8,350 viewsSolve:
_{9}C_{3} = 
Binomial Theorem
Problem 2 7,596 viewsExpand:
(2x − 3)⁴ 
Binomial Theorem
Problem 3 6,767 viewsFind the 4^{th} term of
(x² − 2y)^{10} 
Binomial Theorem
Problem 4 1,597 views 
Binomial Theorem
Problem 5 1,744 views 
Binomial Theorem
Problem 6 1,426 views 
Binomial Theorem
Problem 7 1,399 views 
Binomial Theorem
Problem 8 1,325 views
Comments (0)
Please Sign in or Sign up to add your comment.
·
Delete