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 Concerning combinatorics (Posted on 2011-10-04)
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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 1 step further | Comment 6 of 7 |
Well yes, the rows do not need to be the same.  Jer uses as an example:

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.```

This is the same situation, except not counting from the halfway points.

Count the number of ways to get from (0,0) to (5,2) in 7 (i.e., 3+4) steps.  It is C(7,2) = 21 steps.

Alternatively, consider the set of points which are 3 steps from (0,0) and 4 steps from (7,2).

There are 3 ways to get there via (1,2).  3 ways to get to (1,2) times 1 way to get to (5,2) from there.

There are 12 ways to get there via (2,1).  3 ways to get to (2,1) times 4 ways to get to (5,2) from there.

And there are 6 ways to get there via (0,3).  1 way to get to (0,3) times 6 ways to get to (5,2) from there.

So, 3*1 + 3*4 + 6*1 = C(7,2) = 21

 Posted by Steve Herman on 2011-10-06 00:45:18

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