Eights rooks are placed randomly on a chess board. What is the probability that none of the rooks can capture any of the other r
1 answer:
Answer:
The probability is
Step-by-step explanation:
We can divide the amount of favourable cases by the total amount of cases.
The total amount of cases is the total amount of ways to put 8 rooks on a chessboard. Since a chessboard has 64 squares, this number is the combinatorial number of 64 with 8,
For a favourable case, you need one rook on each column, and for each column the correspondent rook should be in a diferent row than the rest of the rooks. A favourable case can be represented by a bijective function with A = {1,2,3,4,5,6,7,8}. f(i) = j represents that the rook located in the column i is located in the row j.
Thus, the total of favourable cases is equal to the total amount of bijective functions between a set of 8 elements. This amount is 8!, because we have 8 possibilities for the first column, 7 for the second one, 6 on the third one, and so on.
We can conclude that the probability for 8 rooks not being able to capture themselves is
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Answer:
C
Step-by-step explanation:
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Answer:
. hole or removable discontinuity at x=2
Step-by-step explanation:
Well generally if you want the simplest form, you factor each the denominator and numerator and then see if you can cancel any of the factors out (because they're in the denominator and numerator)
So let's start by factoring the first equation:
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Multiply a and c
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Factors that add up to -7 and multiply to 12:
Rewrite equation:
Group terms:
Factor out GCF:
Rewrite:
Now let's write out the equation using these factors:
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Here we can factor out the x-2 and the simplified form is:
So we can "technically" define f(2) using the most simplified form, but it's removable discontinuity, so it has a hole as x=2. since it makes (x-2) equal to 0 (2-2) = 0.