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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)

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Some Thoughts Possible explanation | Comment 1 of 7

Number the rows of Pascals Triangle 0 to R and the elements of each row 0 to E. 

I    Each element in the triangle is then described by the formula R!/(E!(R-E)!)

II    The stipulated sum for the Rth row, where the offset is 1, is ((2R)!)/((R-1)! (R+1)!).  Verifying with odd and even rows:

R=2 ((2R)!)/((R-1)! (R+1)!)=0+R!/((R-2)!(R-(R-2))!)*R!/((R-1)!(R-(R-1))!)+ R!/((R-2)!(R-(R-2))!)*R!/((R-1)!(R-(R-1))!)+0 gives the expected integer solution 2.     
R=3 ((2R)!)/((R-1)! (R+1)!)=0+R!/((0)!(R-0)!)*R!/((R-2)!(R-1)!)+R!/((R-2)!(R-1)!)*R!/((R-2)!(R-1)!)+0+R!/((0)!(R-0)!)*R!/((R-2)!(R-1)!) gives the expected integer solution 3.     

III   But in the 2Rth row, each element is (2R)!/((E)!((2R)-(E))!) and by substituting E=(R-1) we can obtain (2R)!/((R-1)!((2R)-(R-1))!).

IV   Since ((2R)-(R-1))! easily cancels to (R+1)!, the two expressions are equivalent.

V Similarly, ((2R)!)/((R-p)! (R+p)!) where p is the offset, holds for all positive values of p (noting that, for p>R, the value is 0).

Edited on October 4, 2011, 2:17 pm
  Posted by broll on 2011-10-04 12:53:20

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