Calculating n!/(k-1)!(n-k+1)! from Binomial Theorem

AI Thread Summary
The discussion focuses on deriving the formula n!/(k-1)!(n-k+1)! from the binomial coefficient \binom{n}{k-1}. Initially, there is confusion about whether the denominator should include n-k-1 instead of n-k+1. Upon further examination, it is clarified that the correct expression indeed includes n-k+1, aligning with the binomial theorem's application. The realization comes from recognizing how the binomial theorem can be utilized for different purposes. This highlights the importance of careful attention to the terms in mathematical formulas.
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1. How do you get n!/(k-1)!(n-k+1)! from <br /> \begin{pmatrix}<br /> n\\k-1<br /> \end{pmatrix}<br />

I thought it would be n!/(k-1)!(n-k-1)! where the n-k+1 on the bottom of the fraction would be a n-k-1 instead. I don't understand why there is a "+1" wouldn't you just replace k with k-1 in the binomial formula?
 
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\binom{n}{k-1}=\frac {n!}{(k-1)!(n-(k-1))!}=\frac{n!}{(k-1)!(n-k+1))!}
 
I just got it after i typed that out. I can't believe i didnt notice that
 
Binomial theorem use for different purposes. The above person share write formula.
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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