I'm having problems wrapping my head around this...so I want to post some questions out of this Introductory Combinatorics book and what I believe to be the solutions:(adsbygoogle = window.adsbygoogle || []).push({});

1.) There are 6 rooks placed on a 6 by 6 chessboard. How many ways if there are 2 red and 4 blue?

I got 6!*\binom{6}{4} as there are 6! places to put it and 6 choose 4 ways to pain the rooks blue.

2.) A classroom has 2 rows of 8 seats. There are 14 students, 5 of whom always sit in the front row and 4 whom always sit in the back.

Does this work out in roughly the same way?

\binom{9}{3}*3!+\binom{5}{4}*4!

there are 9 people left who won't sit up front, therefore leaving 3 spots to choose for the nine. In the back there will be 14-5 (that sit up front)-4(the ones who will always be in back). So there will be 5 choose 4*4!

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# Problems with the Multiplication Principle (combinatorics)

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