Re: Combinatorica questions!!!
- To: mathgroup at smc.vnet.net
- Subject: [mg20552] Re: Combinatorica questions!!!
- From: Ed McBride <emcbride at wybron.com>
- Date: Sat, 30 Oct 1999 00:14:03 -0400
- Organization: Wybron, Inc.
- References: <firstname.lastname@example.org>
- Sender: owner-wri-mathgroup at wolfram.com
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.