Do you know what nC5 and nC4 represent?

  • Thread starter Thread starter dancing_math
  • Start date Start date
AI Thread Summary
The expression nC5/nC4 can be simplified using the factorial formula n!/(n-r)!r!. The initial attempt leads to the expression 1/(n-5)!5! x (n-4)!4!, which can be further simplified. The simplification involves reducing 4!/5! and (n-4)!/(n-5)!. Participants discuss possible answers, including A. 1/5(n-5), B. (n-4)/5, and C. 5(n-5). The discussion emphasizes the importance of correctly applying factorial simplifications to reach the final answer.
dancing_math
Messages
1
Reaction score
0

Homework Statement



The expression nC5/nC4 can be simplified to:

Homework Equations



I'm assuming that I'm to use n!/(n-r)!r!

The Attempt at a Solution



I attempted to do this:
n!/(n-5)!5! / n!/(n-4)!4!
n!/(n-5)!5! x (n-4)!4!/n!
1/(n-5)!5! x (n-4)!4!
And I'm stuck after this point. The possible answers are:
A. 1/5(n-5)
B. (n-4)/5
C. 5(n-5)
D. none of the above.
 
Physics news on Phys.org
dancing_math said:

Homework Statement



The expression nC5/nC4 can be simplified to:

Homework Equations



I'm assuming that I'm to use n!/(n-r)!r!

The Attempt at a Solution



I attempted to do this:
n!/(n-5)!5! / n!/(n-4)!4!
n!/(n-5)!5! x (n-4)!4!/n!
1/(n-5)!5! x (n-4)!4!
And I'm stuck after this point. The possible answers are:
A. 1/5(n-5)
B. (n-4)/5
C. 5(n-5)
D. none of the above.
Hello dancing_math. Welcome to PF !


1/(n-5)!5! x (n-4)!4! is equivalent to \displaystyle \frac{(n-4)!\,4!}{(n-5)!\,5!}\ .

You should be able to further simplify \displaystyle \frac{4!}{5!} and \displaystyle \frac{(n-4)!}{(n-5)!}\ .
 
I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...
Essentially I just have this problem that I'm stuck on, on a sheet about complex numbers: Show that, for ##|r|<1,## $$1+r\cos(x)+r^2\cos(2x)+r^3\cos(3x)...=\frac{1-r\cos(x)}{1-2r\cos(x)+r^2}$$ My first thought was to express it as a geometric series, where the real part of the sum of the series would be the series you see above: $$1+re^{ix}+r^2e^{2ix}+r^3e^{3ix}...$$ The sum of this series is just: $$\frac{(re^{ix})^n-1}{re^{ix} - 1}$$ I'm having some trouble trying to figure out what to...
Back
Top