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Concerning combinatorics (Posted on 2011-10-04) Difficulty: 3 of 5
Take a row of Pascals Triangle. Write a second copy of the row below it but with an offset. Multiply each pair of numbers and then sum these products. (A missing number is considered to be zero.)

The result is number further down the Triangle.

Can you explain or prove the result?

Example with the 4th row and offset by 1:

1    4    6    4    1    0
0    1    4    6    4    1
0 +  4 + 24 + 24 +  4 +  0 = 56 which is in row 8.

No Solution Yet Submitted by Jer    
Rating: 4.5000 (2 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Result explaned - No Proof | Comment 2 of 7 |

If n is the row in the Pascal triangle
( n >= 0 ) and k is the offset
( 0 <= k <= n ), then
    n-k
  ______
  \
   \   C(n,i)*C(n,i+k)  =  C(2n,n-k)
   /
  /
  ------
    i=0
where 
                     m!
       C(m,n) = -----------
                 n!*(m-n)!
The sum lies in row 2n at element n-k
( also at element n+k ).
I have checked this out with a computer
program for rows n=0 to n=8 and offsets
k=0 to k=n.
Note: For n=0 the result does not lie
      further down the triangle.
I am still trying to prove it. Tried
mathematical induction, but that doesn't
seem to work.

  Posted by Bractals on 2011-10-04 12:56:56
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