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aleksley [76]
3 years ago
11

12 + 36 /3 × 4 + 300

Mathematics
1 answer:
kiruha [24]3 years ago
6 0
360 is the answer
1. 36/3 = 12
2. 12 x 4 = 48
3. 12 + 48 + 300 = 360
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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
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Answer:

The probability is \frac{56!}{64!}

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, 64 \choose 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  f : A \rightarrow A , 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

\frac{8!}{64 \choose 8} = \frac{8!}{\frac{64!}{8!56!}} = \frac{56!}{64!}

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After you've calculated a least common multiple, always check to be sure your answer can be divided evenly by both numbers. Find the LCM of these sets of numbers. Multiply each factor the greatest number of times it occurs in any of the numbers. 9 has two 3's, and 21 has one 7, so we multiply 3 two times, and 7 once.

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Step-by-step explanation:

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Step-by-step explanation:

3:4 itself.

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