The are 40320 ways in which the 5 indistinguishable rooks be can be placed on an 8-by-8 chess- board so that no rook can attack another and neither the first row nor the first column is empty
<h3 /><h3>What involves the
rook polynomial? </h3>
The rook polynomial as a generalization of the rooks problem
Indeed, its result is that 8 non-attacking rooks can be arranged on an 8 × 8 chessboard in r8.
Hence, 8! = 40320 ways.
Therefore, there are 40320 ways in which the 5 indistinguishable rooks be can be placed on an 8-by-8 chess- board so that no rook can attack another and neither the first row nor the first column is empty.
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Answer:
113°
Step-by-step explanation:
1) according to the condition m(∠B)=m(∠D)=67°;
2) m(∠C)=180°-m(∠B)=180°-m(∠D);
m(∠C)=180-67=113°.
Answer:
1/6 or 16%
Step-by-step explanation:
There is a total of 150 students 87 boys and 63 girls
We see their is 25 boys with red hair
So we divide that amount by the total 25/150
We get 16% or 1/6
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Answer:
36
Step-by-step explanation:
It would only be 36 if they split the amount though
Answer:
21 box of books can be brought up by the delivery person at one time
Step-by-step explanation:
Here, we want to know the number of box of books that the delivery person could bring up at one time.
Let the number of boxes be x , so the total mass of the boxes that could come up at a time will be x * 40 = 40x lb
Let’s add this to the mass of the delivery person = 150 lb
So the total mass going inside the lift would be 150 + 40x
So we have to equate this to the maximum capacity of the lift;
150 + 40x = 1020
40x = 1020 - 150
40x = 870
x = 870/40
x = 21.75
Now since we cannot have fractional boxes, the number of boxes that could come into the lift without exceeding the maximum capacity of the lift is 21
