# Thread: (n choose k) How can I work this out?

1. ## (n choose k) How can I work this out?

I know what n choose k means but usually I get given both the values.

2. ## re: (n choose k) How can I work this out?

Originally Posted by uperkurk

I know what n choose k means but usually I get given both the values.
This means $8!(N-8)!=4!(N-4)!$

From that we get, $8\cdot 7\cdot 6\cdot 5=(N-4)(N-5)(N-6)(N-7)$

3. ## re: (n choose k) How can I work this out?

I don't get it... what value is N?

But how do I compute that using my calculator? N still does not have a value, you're saying 8x7x6x5 = (N - 4) but what is N?

4. ## re: (n choose k) How can I work this out?

Originally Posted by uperkurk

I know what n choose k means <--- ähemm?
but usually I get given both the values.
Originally Posted by uperkurk
I don't get it... what value is N?

But how do I compute that using my calculator? <--- you don't have to use a calculator!
N still does not have a value, you're saying 8x7x6x5 = (N - 4) but what is N?
1. Definbition: $\displaystyle{{n \choose k} = \frac{n!}{k! \cdot (n-k)!}}$

$\displaystyle{\frac{n!}{8! \cdot (n-8)!} = \frac{n!}{4! \cdot (n-4)!}}$

2. Divide through by n!. Cross-multiply and you'll get:

$\displaystyle{\frac{(n-4)!}{(n-8)!} = \frac{8!}{4!}~\implies~\frac{(n-4) \cdot (n-5) \cdot (n-6) \cdot (n-7) \cdot (n-8) \cdot (n-9) \cdot ... \cdot 1}{(n-8) \cdot (n-9) \cdot ... \cdot 1} = \frac{8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot ... \cdot 1}{4 \cdot 3 \cdot ... \cdot 1}}$

3. Cancel equal factors and you'll get:

$\displaystyle{(n-4) \cdot (n-5) \cdot (n-6) \cdot (n-7) = 8 \cdot 7 \cdot 6 \cdot 5$

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 $\displaystyle{n \in \mathbb{N}}$ this solution isn't valid.
You'll get the solution by comparison of the factors. And that's your turn.

5. ## re: (n choose k) How can I work this out?

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 :/

6. ## re: (n choose k) How can I work this out?

Originally Posted by uperkurk
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 <--- you use the equal sign as a command to evaluate a constant term, but that's not solving an equation for a variable

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:

$\displaystyle{{n \choose k} = \frac{n!}{k! \cdot (n-k)!}}$

with the values of your 1st example:

$\displaystyle{{10 \choose 7} = \frac{10!}{7! \cdot (10-7)!}} = (10! / 7!) / 3!$

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.

7. ## re: (n choose k) How can I work this out?

Originally Posted by uperkurk
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 :/
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 $\binom{N}{9}=\binom{N}{6}$, we look at it and see at once $N=15$.
To understand why, one has to understand how $\binom{N}{k}=\frac{N!}{k!(N-k)!}$ works.

It is easy to prove that $\binom{N}{k}=\binom{N}{N-k}$

So if $\binom{N}{k}=\binom{N}{j}$ then $k=N-j$ or $N=k+j$.

8. ## re: (n choose k) How can I work this out?

Originally Posted by Plato
It is easy to prove that $\binom{N}{k}=\binom{N}{N-k}$

So if $\binom{N}{k}=\binom{N}{j}$ then $k=N-j$ or $N=k+j$.
Strictly speaking, $\binom{N}{k}=\binom{N}{N-k}$ is not sufficient to conclude that $\binom{N}{k}=\binom{N}{j}$ implies $N=k+j$. We could also use the fact that for a given N, the sequence $\binom{N}{n}$ for n = 0, ..., N goes first up and then down, so that for each M there exist at most two n's such that $\binom{N}{n}=M$.

Originally Posted by Plato
From that we get, $8\cdot 7\cdot 6\cdot 5=(N-4)(N-5)(N-6)(N-7)$
Originally Posted by uperkurk
I don't get it... what value is N?

But how do I compute that using my calculator? N still does not have a value, you're saying 8x7x6x5 = (N - 4) but what is N?
To repeat, $8\cdot 7\cdot 6\cdot 5=(N-4)(N-5)(N-6)(N-7)$ 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 $N^2 + 3N - 4 = 0$ or $2N + 3 = 5$) 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.

9. ## re: (n choose k) How can I work this out?

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

$\binom{10}{7}=\frac{10!}{7!(10-7)!}$

But if I just said

$\binom{N}{7}=\binom{N}{9}$

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

10. ## re: (n choose k) How can I work this out?

Originally Posted by uperkurk
But if I just said

$\binom{N}{7}=\binom{N}{9}$
The answer is $N=7+9=16$
Throw your calculator away or at least put in a drawer.
You don't need it for these problems.
You have a brain.

11. ## re: (n choose k) How can I work this out?

What so literally that's all you do? Is add the two k numbers and you have your N number?

$\binom{N}{3}=\binom{N}{7}$

$N = 10?$

12. ## re: (n choose k) How can I work this out?

Originally Posted by uperkurk
What so literally that's all you do? Is add the two k numbers and you have your N number?
For equations of these particular form, yes. This does not work, for example, for $\binom{N}{3}=\binom{N+1}{7}$. Also, it helps to understand why the sum of the two bottom numbers is the solution. This was explained in posts #7 and #8.

Originally Posted by uperkurk
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
It's the multiplication sign.

Also, the word is "formula," not "formular."

13. ## re: (n choose k) How can I work this out?

In the exam I'll most likely have to provide some working out, I'll just say what you said

$k = N - j$ or $N = k + j$