Binomial Theorem

[03myersd]

Just started doing binomial theorem and I am having a bit of a problem with my maths. The question is:

(n) = 10
(3)

(Basically one big bracket) and I need to get n.

This is what I have so far:

(n) = 10
(3)

n! / ((n-3)!*3!) = 10

n! / (n-3)! = 10 * 3!

n! / (n-3)! = 60

n(n-1)(n-2) = 60

n(n² - 2n - n + 3) = 6-

n(n² - 3n + 3) = 60

n³ - 3n² + 3n = 60

n³ - 3n² + 3n - 60 = 0

Then using a graphical calculator to get a root I am getting 4.9 (roughly) which I know must be wrong as it is wrong in the back of the book. What am I doing wrong or forgetting?

[NickThissen]

2 * 1 = 2, not 3

(n-1)(n-2) = n^2 - 3n + 2

[03myersd]

I knew that.... was just a test for everyone.... *cough*

Thanks Nick!!!


The funny thing is that 2 people have checked over my working for me and they both missed it.

[anhn]

n(n-1)(n-2) = 60
n(n² - 2n - n + 2) = 60
n(n² - 3n + 2) = 60
n³ - 3n² + 2n = 60
n³ - 3n² + 2n - 60 = 0
n = 5

Edit: my bad, it's too late

[jemidiah]

You can also look back at pascal's triangle and use the fact that its entries are actually binomial coefficients. Since the only 10 that appears in the third "column" [imagine the triangle to be a right triangle instead of isosceles, with the left-most entries lined up] is in the 5th row, you know that 5 choose 3 = 10, so n=5.

[krtxmrtz]

I wonder if you're really supposed to go about solving 3rd degree equations even with a calculator, I'd say not. Therefore jemidiah's suggestion about Pascal's triangle is probably the way to go. Quite often what it's all about is thinking of some clever solution rather than fooling around with correct yet cumbersome algebraic calculations.

[03myersd]

We either use graphical calculators to get a root, or use a trial and error method when synthetically dividing to get the root.

[anhn]

n(n-1)(n-2) = 60
(n-2)(n-1)n = 2²*3*5
(n-2)(n-1)n = 3*4*5
n = 5