Solving for A Variable in Combination

AI Thread Summary
To solve the equation _5C_n = 10, the user starts with the combination formula, simplifying it to \frac{120}{n!(5-n)!}=10. They express frustration over the guess-and-check method for finding n, while noting that there are likely two solutions within the range of 0 to 5. The discussion also touches on the difficulty of using the Online LaTeX Equation Editor for typing math equations. Ultimately, the user seeks a more efficient method for solving combinations and better tools for formatting equations.
lacrotix
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Homework Statement



Solve for _5C_n = 10

The Attempt at a Solution



_5C_n = 10

\frac{5!}{n!(5-n)!}=10

\frac{120}{n!(5-n)!}=10

I can't go on from there. I remember that my teacher told me that for combinations and permutations, if the variable is the second number (or number to the right of the big letter), that you have to try to simply the best you can logically, and then guess based on hunches. I also remember him also saying that looking at the equation will help you make an "informed hunch."

The problem is, I don't think I simplified my problem right to make these guesses. I guess I can always guess and check straight from the beginning, but I think there should be an easier way than brute force guessing and checking. Especially if the equation is something like:

_2_5_0C_n = 2573000

Any ideas?

Also, is there an easier way of typing in math equations for homework help. It is so difficult for me to guess with the Online LaTeX Equation Editor. Yuck!
 
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\frac{120}{n!(5-n)!}=10

I don't see any way besides guess-and-check at the moment. It's pretty obvious that
n!(5-n)!=12
and assuming that
0 \le n \le 5
there won't be many guesses you'll have to make until you find your answer. (BTW, there are two answers.)

lacrotix said:
Also, is there an easier way of typing in math equations for homework help. It is so difficult for me to guess with the Online LaTeX Equation Editor. Yuck!
You can find guides easily enough that show how to type equations in LaTeX. Just keep practicing and it won't be difficult after a while.
 
Thanks eumyang.

That description and logic-checking helps a whole lot.
 
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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