Proving a combinations identity (1 Viewer)

[SunnyScience]

From 1997 HSC Q5 b

Prove:
nCr + nCr+1 = n+1Cr+1

I don't understand the books solution. Can anyone show me how to do it (with full working), please?

Thanks

 

[Drongoski]

From 1997 HSC Q5 b

Prove:
nCr + nCr+1 = n+1Cr+1

I don't understand the books solution. Can anyone show me how to do it (with full working), please?

Thanks

What part of the book's solution (please post this also) you don't understand?

 

[SunnyScience]

(nCr) + (nCr+1)
= n!/r!(n-r)! + n!/(r+1)!(n-r-1)!
= n!(r+1) + n!(n-r) / (r+1)!(n-r)!

The last time particularly - rolling it of factorials

ty

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[SunnyScience]

*out

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[SunnyScience]

Bump

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[Drongoski]

From 1997 HSC Q5 b

Prove:
nCr + nCr+1 = n+1Cr+1

I don't understand the books solution. Can anyone show me how to do it (with full working), please?

I'll use equivalent notation.



This says for example:







Sorry - I have to rush off to tutor. Will complete later if necessary.

 

[Carrotsticks]

I'll finish it for you don't worry.

 

[Nooblet94]

I'll use equivalent notation.



This say for example:







Sorry - I have to rush off to tutor. Will complete later if necessary.

Continuing from above:
<a href="http://www.codecogs.com/eqnedit.php?latex=\\ =\frac{n!(r@plus;1)}{(r@plus;1)!(n-r)!}@plus;\frac{n!(n-r)}{(r@plus;1)!(n-r)!}\\ =\frac{n!(r@plus;1@plus;n-r)}{(r@plus;1)!(n-r)!}\\ =\frac{n!(n@plus;1)}{(r@plus;1)!(n-r)!}\\ =\frac{(n@plus;1)!}{(r@plus;1)!(n-r)!}\\ =\frac{(n@plus;1)!}{(r@plus;1)![(n@plus;1)-(r@plus;1)]!}\\ =RHS" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\\ =\frac{n!(r+1)}{(r+1)!(n-r)!}+\frac{n!(n-r)}{(r+1)!(n-r)!}\\ =\frac{n!(r+1+n-r)}{(r+1)!(n-r)!}\\ =\frac{n!(n+1)}{(r+1)!(n-r)!}\\ =\frac{(n+1)!}{(r+1)!(n-r)!}\\ =\frac{(n+1)!}{(r+1)![(n+1)-(r+1)]!}\\ =RHS" title="\\ =\frac{n!(r+1)}{(r+1)!(n-r)!}+\frac{n!(n-r)}{(r+1)!(n-r)!}\\ =\frac{n!(r+1+n-r)}{(r+1)!(n-r)!}\\ =\frac{n!(n+1)}{(r+1)!(n-r)!}\\ =\frac{(n+1)!}{(r+1)!(n-r)!}\\ =\frac{(n+1)!}{(r+1)![(n+1)-(r+1)]!}\\ =RHS" /></a>

 

[Drongoski]

Thanks Nooblet for finishing it off.

 

[SunnyScience]

Thanks guys

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[Trebla]

Note that this identity is precisely what you use to form Pascal's triangle i.e. add two adjacent numbers in a given row to obtain the number 'below' them in the next row.