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Re: Combinatorica questions!!!

  • To: mathgroup at
  • Subject: [mg20552] Re: Combinatorica questions!!!
  • From: Ed McBride <emcbride at>
  • Date: Sat, 30 Oct 1999 00:14:03 -0400
  • Organization: Wybron, Inc.
  • References: <7v6726$>
  • Sender: owner-wri-mathgroup at

Keren Edwards wrote:
> Hi all!!
> 2 different questions:
> 1.    how many ways does a castle have to reach from the bottom left side
> corner
>        of a chess board to the upper right corner of the board if he can
> move right
>        and up only?
> 2.     you have 8 red identical balls, 9 purple identical balls and 7 white
> identical ones.
>         a.  How many ways can you choose 10 balls with no matter to the
> order of the balls?
>         b.  How many ways can you choose 10 balls with no matter to the
> order of the balls, if each color must
>               be chosen once at least?
> Many thanx.

1.  Consider diagonals of squares, each running from the upper leftt
direction to the lower right direction, sort of perpendicular to the
general direction of the motion of the castle.  The numbers of ways a
castle can reach the squares in one of these diagonal rows is a row in
Pascal's triangle.  So the numbers for the squares in the main diagonal
are 1, 7, 21, 35, 35, 21, 7, 1.  The problem is symmetric, so if there
are n ways to reach a main-diagonal square from the lower left corner,
there are n ways to reach the upper left corner from that same
main-diagonal square.  The final answer is the sum of the squares of the
eight numbers listed above.  Which, I think, is 3432.  As an aside, note
that this answer must also be the number of ways to reach the center
square in the main diagonal of a 16 by 16 chessboard.  So, we end up
with a general formula, with little or no practical value as far as I
can tell:  [Sum of squares of Comb(n things, i at a time), i = 0, n]
must equal [Comb (2n things, n at a time)].

Ed mcBride, P.E.

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