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  1. How many ways are there of placing $5$ non-taking rooks on $5 \times 5$ board?
  2. How many ways if none lie on the main diagonal?
  3. How many ways if exactly one lies on the main diagonal?

My solution:

The first is $5!$, this was easy.

I don't actually have an idea for the second and third :( Can someone help me, thanks :)

Eman Yalpsid
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Johnathan
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  • Consider all possibilities and take out $\cup A_i,$ where $A_i$ are the possibilities with a rook on the cell $(i,i)$ in the diagonal. Can you count $|A_i|$? What about $|A_{i}\cap A_j|$? – Phicar Mar 19 '17 at 16:32
  • Take a look at my answer here, although it's not identical, you should be able to see that such arrangements for part 2 can be reinterpreted as derangements and for part 3 as permutations with 1 fixed point. Part 1 I actually answer directly (note that this question probably assumes rooks are identical - just my intuition speaking here but the only difference is a factor of $5!$). – N. Shales Mar 20 '17 at 02:33

1 Answers1

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First part of the question: There are 25 places for the first rook for the next rook there are (25-9)=16 places. For the third rook there are (16-7)=9 placesthe b similarly for the fourth and fifth rook there are 4 and 1 place respectively. Therefore the board can be set up in 25×16×9×4×1=14400 ways.

Second part of the question: First we have to consider in how many ways can we setup the board if we each rook lies on the diagonal. If we consider that each rook lies in the diagonal then first rook can be placed in 5 places second rook in 4 places and so on. Therefore board can be setup in 5! ways if each rook lies on the main diagonal so if none of the rooks lie on the same diagonal then the board can be setup in (14400-5!=14400-120=14280)

For the third part: It is exactly like the first except for the first rook there are 5 places and for rest of rooks same so the board can be setup in (5×16×9×4×1=2880) ways.

Dev Rajput
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