# factorial question

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#1
Question:

Simplify

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

my attempt;

I cancel the top and bottom by (r-2)!

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

I cancel out the top and bottom by (r-1)!

I cancel out the top and bottom by (r-1)!

i cancel out the top and bottom by (n-r+1)!

0
2 years ago
#2
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

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

my attempt;

I cancel the top and bottom by (r-2)!

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

I cancel out the top and bottom by (r-1)!

I cancel out the top and bottom by (r-1)!

i cancel out the top and bottom by (n-r+1)!

Last edited by mqb2766; 2 years ago
0
2 years ago
#3
(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
0
2 years ago
#4
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 ?
1
2 years ago
#5
(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
0
2 years ago
#6
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
0
#7
(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
0
2 years ago
#8
Sure, but the 2nd approach with factorization must be easier?

(Original post by bigmansouf)
thanks but i have not learnt about Pascal's triangle
0
2 years ago
#9
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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