I know what n choose k means but usually I get given both the values.
1. Definbition: $\displaystyle \displaystyle{{n \choose k} = \frac{n!}{k! \cdot (n-k)!}}$
With your equation you get:
$\displaystyle \displaystyle{\frac{n!}{8! \cdot (n-8)!} = \frac{n!}{4! \cdot (n-4)!}}$
2. Divide through by n!. Cross-multiply and you'll get:
$\displaystyle \displaystyle{\frac{(n-4)!}{(n-8)!} = \frac{8!}{4!}~\implies~\frac{(n-4) \cdot (n-5) \cdot (n-6) \cdot (n-7) \cdot (n-8) \cdot (n-9) \cdot ... \cdot 1}{(n-8) \cdot (n-9) \cdot ... \cdot 1} = \frac{8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot ... \cdot 1}{4 \cdot 3 \cdot ... \cdot 1}}$
3. Cancel equal factors and you'll get:
$\displaystyle \displaystyle{(n-4) \cdot (n-5) \cdot (n-6) \cdot (n-7) = 8 \cdot 7 \cdot 6 \cdot 5$
and that's the equation Plato has posted in his post (and that's what you should have got because you know what n choose k means)
4. This is a quartic equation in n. By first inspection you get n = -1 as a solution. But since $\displaystyle \displaystyle{n \in \mathbb{N}}$ this solution isn't valid.
You'll get the solution by comparison of the factors. And that's your turn.
I've never used that formular when solving these equations. Take for example 10 choose 7. I just do 10! / 7! / 3! and I get 120
and for example 20 choose 5 = 20! / 5! / 15! = 15,504
I don't know how to use that formular to answer these questions :/
There isn't much of a difference between the formula you use and mine:
$\displaystyle \displaystyle{{n \choose k} = \frac{n!}{k! \cdot (n-k)!}}$
with the values of your 1st example:
$\displaystyle \displaystyle{{10 \choose 7} = \frac{10!}{7! \cdot (10-7)!}} = (10! / 7!) / 3!$
Instead of using fixed values your equation (from your 1st post) contains a variable n. So you have to transform the given equation (which is only possible using the general form) such that you are able to determine a value for n. That's what I've done for you. Now go ahead and solve for n.
The reason that I answered the way I did is that I suspected that you may not have understood the principle.
Actually this is a vary simple question.
In $\displaystyle \binom{N}{9}=\binom{N}{6}$, we look at it and see at once $\displaystyle N=15$.
To understand why, one has to understand how $\displaystyle \binom{N}{k}=\frac{N!}{k!(N-k)!}$ works.
It is easy to prove that $\displaystyle \binom{N}{k}=\binom{N}{N-k}$
So if $\displaystyle \binom{N}{k}=\binom{N}{j}$ then $\displaystyle k=N-j$ or $\displaystyle N=k+j$.
Strictly speaking, $\displaystyle \binom{N}{k}=\binom{N}{N-k}$ is not sufficient to conclude that $\displaystyle \binom{N}{k}=\binom{N}{j}$ implies $\displaystyle N=k+j$. We could also use the fact that for a given N, the sequence $\displaystyle \binom{N}{n}$ for n = 0, ..., N goes first up and then down, so that for each M there exist at most two n's such that $\displaystyle \binom{N}{n}=M$.
To repeat, $\displaystyle 8\cdot 7\cdot 6\cdot 5=(N-4)(N-5)(N-6)(N-7)$ is not an equation of the form N = ..., which allows calculating N immediately. Rather, it is a constraint on N (of the same sort as $\displaystyle N^2 + 3N - 4 = 0$ or $\displaystyle 2N + 3 = 5$) and has to be solved first. It is an equation of fourth degree, and two solutions can be easily guessed (earboth noted that N = -1 is a solutions). The two other solutions turn out to be complex numbers, so they are irrelevant for the problem. However, Plato's method at the top of the post is easier.
Ok so here I see how you have come to 15, just add the two numbers together, I'm not all that familiar with mathematic notation, what does the . mean between 8.7.6.5 and the thing that throws me off is you're saying N - 5 or whatever
So I can put this into the formular which would look like this
$\displaystyle \binom{10}{7}=\frac{10!}{7!(10-7)!}$
But if I just said
$\displaystyle \binom{N}{7}=\binom{N}{9}$
I don't understand your formular, I'm really bad at reading formulars
P.S That formular still does not make sense to me... in my calculator I type, 10! divided by 7! divided by 10 - 7!
I don't think I'm reading the formular correctly
For equations of these particular form, yes. This does not work, for example, for $\displaystyle \binom{N}{3}=\binom{N+1}{7}$. Also, it helps to understand why the sum of the two bottom numbers is the solution. This was explained in posts #7 and #8.
It's the multiplication sign.
Also, the word is "formula," not "formular."