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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.)
Some Thoughts re: 1 step further. Not commutative. Comment 7 of 7 |
(In reply to 1 step further. by Jer)


Here is the 3rd times the 4th row with an offset of 1:
1    3    3    1    0    0
0    1    4    6    4    1
0 +  3 + 12 +  6 +  0 +  0 = 21 which is in row 7.
How about 4th times 3rd row with offset of 1:
1    4    6    4    1   
0    1    3    3    1
0 +  4 + 18 + 12 +  1  = 35 also in row 7.

  Posted by Bractals on 2011-10-07 07:30:00
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