factorial question

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)!}

Reply 1

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

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)!}

(edited 4 years ago)

Reply 2

T!gger34

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 :redface:

Edit - just done it. Nice one and simple!!!!!!!

(edited 4 years ago)

Reply 3

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.

Original post by dextrous63

How does one use Pascal's triangle in this manner? This is a new one to me :redface:

Are you referring to the symmetries of nCr ?

Reply 4

T!gger34

Original post by mqb2766

Yep. Edited my post and have done it. Just need to hop down a couple of lines :wink:

Very elegant and a new technique for me.. Thanks

Reply 5

mqb2766

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 :wink:

Very elegant and a new technique for me.. Thanks

Reply 6

bigmansouf

Original post by mqb2766

thanks but i have not learnt about Pascal's triangle

Reply 7

mqb2766

Sure, but the 2nd approach with factorization must be easier?

Original post by bigmansouf

thanks but i have not learnt about Pascal's triangle

Reply 8

username3555092

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 ?