Question

In
how many ways can 2 red and four blue rooks be placed on am 8 by 8
board so that no 2 rooks can attack one another ?

A roller coaster has 5 cars containing 4 seats 2 in frony and
2 in the back, there are 20 people ready for a ride, how many ways
can the ride begin? what if a certain 2 people want to seat in
different cars?

Answer #1

1st question)

note that there cannot be more than one rook in any column or row in order to avoid any rook attacking any other rooks. the placement can be done to two stages.

stage 1 = place 2 red rooks. choose 2 rows out of 8 rows and there are (8 2) ways. then for each place the two rooks in the two rows chosen and there are 8.7 of choices. thus, there are (8 2).8.7 ways to place 2 red rooks.

stage 2 = for each placement of the two red rooks, choose 4 rows out the six remaining 6 rows to place the four blue rooks. there are ( 6 4) ways. then place the four rooks in the remaining 6 columns for each choice of the four rows. there 6.5.4.3 ways to do so. thus, there are (6 4).6.5.4.3 ways to place the four blue rooks in the remaining board

by multiplication principle there are ways to place the six rooks on the board

so that no two rooks attack one another

= **8467200**

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there is no constraint?
first urn has at least 1 red ball and 2 blue balls?
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principle)

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