Arrangement of Counters-II

how many different ways may 4 4 counters be placed on the squares of an 10 × 5 10 × 5 grid so that no two counters are in the same row or in the same column?

Hint : - You may decide to solve this first.


The answer is 25200.

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1 solution

There are four counters. Let's put it piece by piece on the board. The first piece can be placed anywhere on the board; thus we have ( 5 ) ( 10 ) = 50 (5)(10)=50 ways to do that. Now for the second counter, it can't be on the same row and column of the first counter, so we have ( 4 ) ( 9 ) = 36 (4)(9)=36 ways to do that. The third counter can't be on the same row and column of the first and second counter, so we have ( 3 ) ( 8 ) = 24 (3)(8)=24 ways to do that. The fourth counter can't be on the same row and column of first, second and third counter, so we have ( 2 ) ( 7 ) = 14 (2)(7)=14 ways to do that.

total number of ways = ( 50 ) ( 36 ) ( 24 ) ( 14 ) 4 ! = 25200 \frac{(50)(36)(24)(14)}{4!}=25200

Note that we divided it by 4 ! 4! because the order in which the counters are placed is not important.

nice! That was my method as well.

Freddie Hand - 4 years, 5 months ago

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