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Simplify \(\frac{\binom{n}{k}}{\binom{n}{k - 1}}\)

 Apr 14, 2020
 #1
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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
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See detailed answer here:  https://web2.0calc.com/questions/help_73560

 Apr 14, 2020
 #3
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The problem is I don't understand his solution.

 Apr 14, 2020
 #4
avatar+109492 
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Then you need to do some study on your own.

This is very basic to this topic.

Melody  Apr 14, 2020
 #5
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\(\begin{pmatrix} n\\ k \end{pmatrix}\;means \;\; \frac{n!}{k!(n-k)!}\)

 

You need to know this!

Melody  Apr 14, 2020
 #6
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Thanks for the information!

 Apr 15, 2020
 #7
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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
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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
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Why is it +1 Instead of -1?

 

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

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 Apr 15, 2020
 #9
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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
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Oh, brainfart, yeah thanks!

 Apr 16, 2020
 #12
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So I have one last question, why is this true, \((k-1)!=k!/k\)

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 Apr 16, 2020
 #13
avatar+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
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Thanks Melody!

 Apr 17, 2020

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