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# Binomial Theorem

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Simplify $$\frac{\binom{n}{k}}{\binom{n}{k - 1}}$$

Apr 14, 2020

#1
+1

To the person answering this, can I get an explanation for each step, because I want to learn how to solve this problem and future problems that are like this one. Thanks!

Apr 14, 2020
#2
+1

Apr 14, 2020
#3
+1

The problem is I don't understand his solution.

Apr 14, 2020
#4
+109492
+1

Then you need to do some study on your own.

This is very basic to this topic.

Melody  Apr 14, 2020
#5
+109492
+1

$$\begin{pmatrix} n\\ k \end{pmatrix}\;means \;\; \frac{n!}{k!(n-k)!}$$

You need to know this!

Melody  Apr 14, 2020
#6
+1

Thanks for the information!

Apr 15, 2020
#7
+1

But I still don't understand how N choose K-1 = $$n!/(k-1)!(n-k+1)!$$

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Apr 15, 2020
#10
+109492
+1

I'm am just writing the same as guest here. Thanks guest (below)

$$\begin{pmatrix} n\\ k-1 \end{pmatrix}\\=\frac{n!}{(k-1)!(n-[k-1])!}\\ =\frac{n!}{(k-1)!(n-k+1)!}\\$$

Melody  Apr 15, 2020
#8
+1

Why is it +1 Instead of -1?

$$(n-k+1)!$$

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Apr 15, 2020
#9
+1

Well, have you tried a numerical example? Try this:

4 choose 2 =6. Now, make it like this: 4 choose (3 - 1) . Try substituting 4 choose (3-1) for n choose (k-1) and see how to get the same result as 4 choose 2 = 6. When he shows: n - k + 1, he is actually doing this: (n - (k - 1)). First you would do the innermost brackets:(n - k + 1), because - * - = +. So that 4 choose (3-1) gives you exactly the same result as 4 choose 2 = 6 . Do you understand it now?

Guest Apr 15, 2020
edited by Guest  Apr 15, 2020
#11
+1

Oh, brainfart, yeah thanks!

Apr 16, 2020
#12
+1

So I have one last question, why is this true, $$(k-1)!=k!/k$$

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Apr 16, 2020
#13
+109492
+1

$$\frac{k!}{k}=\frac{1*2*3*\dots (k-2)*(k-1)*k}{k}=1*2*3*\dots (k-2)*(k-1)=(k-1)!$$

Melody  Apr 16, 2020
#14
+1

Thanks Melody!

Apr 17, 2020