bigmansouf
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Question:


Simplify
 \frac{n!}{(n-r)!r!}+ \frac{2 \times n!}{(n-r+1)!(r-1)!} + \frac{n!}{(n-r+2)!(r-2)!}

I would like to know if my answer is correct thank you

my attempt;
 \frac{n!(n-r+1)!(r-1)!(n-r+2)!(r-2)!+2 \times n!(n-r)!r!(n-r+2)!(r-2)!+n!(n-r)!r!(n-r+1)!(r-1)!}{(n-r)!r!(n-r+1)!(r-1)!(n-r+2)!(r-2)!}

 \frac{n!\left [(n-r+1)!(r-1)!(n-r+2)!(r-2)!+2 \times(n-r)!r!(n-r+2)!(r-2)!+(n-r)!r!(n-r+1)!(r-1)!\right ]}{(n-r)!r!(n-r+1)!(r-1)!(n-r+2)!(r-2)!}

 \frac{n!\left [(n-r+1)!(r-1)!(n-r+2)!+2 \times(n-r)!r!(n-r+2)!+(n-r)!r!(n-r+1)!(r-1)\right ]}{(n-r)!r!(n-r+1)!(r-1)!(n-r+2)!}


 \frac{n!\left [(n-r+1)!(r-1)!(n-r+2)!((r-2)(r-3)(r-4)..(2)(1))+2 \times(n-r)!r!(n-r+2)!((r-2)(r-3)(r-4)..(2)(1))+(n-r)!r!(n-r+1)!((r-1)(r-2)(r-3)(r-4)..(2)(1))!\right ]}{(n-r)!r!(n-r+1)!(r-1)!(n-r+2)!(r-2)!}
I cancel the top and bottom by (r-2)!

 \frac{n!\left [(n-r+1)!(r-1)!(n-r+2)!+2 \times(n-r)!r!(n-r+2)!+(n-r)!r!(n-r+1)!(r-1)\right ]}{(n-r)!r!(n-r+1)!(r-1)!(n-r+2)!}


I cancel out the top and bottom by (n-r) !

 \frac{n!\left [((n-r+1)(n-r)..(2)(1))(r-1)!(n-r+2)!+2 \times ((n-r)(n-r-1)..(2)(1))r!(n-r+2)!+((n-r+1)(n-r-1)...(2)(1))r!(n-r+1)!(r-1) \right ]}{((n-r)(n-r-1)..(2)(1))r!(n-r+1)!(r-1)!(n-r+2)!}

 \frac{n!\left [(n-r+1)(r-1)!(n-r+2)!+2 \times r!(n-r+2)!+r!(n-r+1)!(r-1) \right ]}{r!(n-r+1)!(r-1)!(n-r+2)!}

I cancel out the top and bottom by (r-1)!
 \frac{n!\left [(n-r+1)((r-1)..(2)(1))(n-r+2)!+2 \times (r(r-1)..(2)(1))(n-r+2)!+(r(r-1)..(2)(1))(n-r+1)!(r-1) \right ]}{(r(r-1)..(2)(1))(n-r+1)!(r-1)!(n-r+2)!}

 \frac{n!\left [(n-r+1)((r-1)..(2)(1))(n-r+2)!+2 \times (r(r-1)..(2)(1))(n-r+2)!+(r(r-1)..(2)(1))(n-r+1)!(r-1) \right ]}{r!(n-r+1)!((r-1)..(2)(1)) (n-r+2)!}
I cancel out the top and bottom by (r-1)!

 \frac{n!\left [(n-r+1)(n-r+2)!+2 \times r(n-r+2)!+r(n-r+1)!(r-1) \right ]}{r!(n-r+1)!(n-r+2)!}

i cancel out the top and bottom by (n-r+1)!
 \frac{n!\left [(n-r+1)((n-r+2)(n-r+1)..(2)(1))+2 \times r((n-r+2)(n-r+1)..(2)(1))+r ((n-r+1)(n-r)..(2)(1))(r-1) \right ]}{r!((n-r+1)(n-r)..(2)(1))(n-r+2)!}

\frac{n!\left [(n-r+1)(n-r+2)+2 \times r(n-r+2)+r(r-1) \right ]}{r!(n-r+2)!}

 \frac{n!\left [ (n^2 -2nr + 3n+ r^2 -3r +2 )+(2rn-2r^2+4r)+r^2 -r \right ]}{r!(n-r+2)!}

 \frac{n!\left [ (n^2 + 3n+ 2 \right ]}{r!(n-r+2)!}
 \frac{n!\left [ (n+1)(n+2) \right]}{r!(n-r+2)!}

Answer =  \frac{(n+2)!}{r!(n-r+2)!}
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mqb2766
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I take it you're trying to prove
nCr + 2nC(r-1) + nC(r-2) = (n+2)Cr
You realize that you could do it much easier using Pascal's triangle (an identity for nCr)?

Edit: If you're doing it using fractions and factorials, I'd have factored out the common bit at the start
n!/((n-r)!(r-2)!) * (.....)
Probably make things much simpler

(Original post by bigmansouf)
Question:


Simplify
 \frac{n!}{(n-r)!r!}+ \frac{2 \times n!}{(n-r+1)!(r-1)!} + \frac{n!}{(n-r+2)!(r-2)!}

I would like to know if my answer is correct thank you

my attempt;
 \frac{n!(n-r+1)!(r-1)!(n-r+2)!(r-2)!+2 \times n!(n-r)!r!(n-r+2)!(r-2)!+n!(n-r)!r!(n-r+1)!(r-1)!}{(n-r)!r!(n-r+1)!(r-1)!(n-r+2)!(r-2)!}

 \frac{n!\left [(n-r+1)!(r-1)!(n-r+2)!(r-2)!+2 \times(n-r)!r!(n-r+2)!(r-2)!+(n-r)!r!(n-r+1)!(r-1)!\right ]}{(n-r)!r!(n-r+1)!(r-1)!(n-r+2)!(r-2)!}

 \frac{n!\left [(n-r+1)!(r-1)!(n-r+2)!+2 \times(n-r)!r!(n-r+2)!+(n-r)!r!(n-r+1)!(r-1)\right ]}{(n-r)!r!(n-r+1)!(r-1)!(n-r+2)!}


 \frac{n!\left [(n-r+1)!(r-1)!(n-r+2)!((r-2)(r-3)(r-4)..(2)(1))+2 \times(n-r)!r!(n-r+2)!((r-2)(r-3)(r-4)..(2)(1))+(n-r)!r!(n-r+1)!((r-1)(r-2)(r-3)(r-4)..(2)(1))!\right ]}{(n-r)!r!(n-r+1)!(r-1)!(n-r+2)!(r-2)!}
I cancel the top and bottom by (r-2)!

 \frac{n!\left [(n-r+1)!(r-1)!(n-r+2)!+2 \times(n-r)!r!(n-r+2)!+(n-r)!r!(n-r+1)!(r-1)\right ]}{(n-r)!r!(n-r+1)!(r-1)!(n-r+2)!}


I cancel out the top and bottom by (n-r) !

 \frac{n!\left [((n-r+1)(n-r)..(2)(1))(r-1)!(n-r+2)!+2 \times ((n-r)(n-r-1)..(2)(1))r!(n-r+2)!+((n-r+1)(n-r-1)...(2)(1))r!(n-r+1)!(r-1) \right ]}{((n-r)(n-r-1)..(2)(1))r!(n-r+1)!(r-1)!(n-r+2)!}

 \frac{n!\left [(n-r+1)(r-1)!(n-r+2)!+2 \times r!(n-r+2)!+r!(n-r+1)!(r-1) \right ]}{r!(n-r+1)!(r-1)!(n-r+2)!}

I cancel out the top and bottom by (r-1)!
 \frac{n!\left [(n-r+1)((r-1)..(2)(1))(n-r+2)!+2 \times (r(r-1)..(2)(1))(n-r+2)!+(r(r-1)..(2)(1))(n-r+1)!(r-1) \right ]}{(r(r-1)..(2)(1))(n-r+1)!(r-1)!(n-r+2)!}

 \frac{n!\left [(n-r+1)((r-1)..(2)(1))(n-r+2)!+2 \times (r(r-1)..(2)(1))(n-r+2)!+(r(r-1)..(2)(1))(n-r+1)!(r-1) \right ]}{r!(n-r+1)!((r-1)..(2)(1)) (n-r+2)!}
I cancel out the top and bottom by (r-1)!

 \frac{n!\left [(n-r+1)(n-r+2)!+2 \times r(n-r+2)!+r(n-r+1)!(r-1) \right ]}{r!(n-r+1)!(n-r+2)!}

i cancel out the top and bottom by (n-r+1)!
 \frac{n!\left [(n-r+1)((n-r+2)(n-r+1)..(2)(1))+2 \times r((n-r+2)(n-r+1)..(2)(1))+r ((n-r+1)(n-r)..(2)(1))(r-1) \right ]}{r!((n-r+1)(n-r)..(2)(1))(n-r+2)!}

\frac{n!\left [(n-r+1)(n-r+2)+2 \times r(n-r+2)+r(r-1) \right ]}{r!(n-r+2)!}

 \frac{n!\left [ (n^2 -2nr + 3n+ r^2 -3r +2 )+(2rn-2r^2+4r)+r^2 -r \right ]}{r!(n-r+2)!}

 \frac{n!\left [ (n^2 + 3n+ 2 \right ]}{r!(n-r+2)!}
 \frac{n!\left [ (n+1)(n+2) \right]}{r!(n-r+2)!}

Answer =  \frac{(n+2)!}{r!(n-r+2)!}
Last edited by mqb2766; 2 years ago
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dextrous63
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(Original post by mqb2766)
I take it you're trying to prove
nCr + 2nC(r-1) + nC(r-2) = (n+2)Cr
You realize that you could do it much easier using Pascal's triangle (an identity for nCr)?
How does one use Pascal's triangle in this manner? This is a new one to me

Edit - just done it. Nice one and simple!!!!!!!
Last edited by dextrous63; 2 years ago
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mqb2766
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Obviously, the entries for Pascal's triangle are nCr

Part of the "hint" is from the coefficients on the left hand side
1 2 1
where does the term (n+2)Cr appear in the triangle ...

Shouldn't be too hard to work out and happy to provide a bit more (if necessary) after the OP responds.

(Original post by dextrous63)
How does one use Pascal's triangle in this manner? This is a new one to me
Are you referring to the symmetries of nCr ?
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dextrous63
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(Original post by mqb2766)
Obviously, the entries for Pascal's triangle are nCr

Part of the "hint" is from the coefficients on the left hand side
1 2 1
where does the term (n+2)Cr appear in the triangle ...

Shouldn't be too hard to work out and happy to provide a bit more (if necessary) after the OP responds.
Yep. Edited my post and have done it. Just need to hop down a couple of lines Very elegant and a new technique for me.. Thanks
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mqb2766
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Yup, triangles within triangles. It would have kept the ancient Greeks happy.

(Original post by dextrous63)
Yep. Edited my post and have done it. Just need to hop down a couple of lines Very elegant and a new technique for me.. Thanks
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bigmansouf
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(Original post by mqb2766)
I take it you're trying to prove
nCr + 2nC(r-1) + nC(r-2) = (n+2)Cr
You realize that you could do it much easier using Pascal's triangle (an identity for nCr)?

Edit: If you're doing it using fractions and factorials, I'd have factored out the common bit at the start
n!/((n-r)!(r-2)!) * (.....)
Probably make things much simpler
thanks but i have not learnt about Pascal's triangle
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mqb2766
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Sure, but the 2nd approach with factorization must be easier?

(Original post by bigmansouf)
thanks but i have not learnt about Pascal's triangle
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username3555092
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Another approach.

Let's say I have n+2 coins marked 1, 2, 3, ..., n+1, n+2 on my table. How many ways are there to choose r coins? Obviously (n+2)Cr.

Now I separate these coins into two sets: one set with coins 1, 2, 3, ..., n-1, n and another set with coins n+1, n+2.

I still want to get r coins. I can:

(i) Pick all r coins from the first set and 0 coins from the second set. How many ways can I do this?
(ii) Pick r-1 coins from the first set and 1 from the second set. How many ways can I do this?
(iii) Pick r-2 coins from the first set and 2 from the second set. How many ways can I do this?

Why is the sum of (i), (ii) and (iii) the same as (n+2)Cr ?
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