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Alenkasestr [34]
3 years ago
6

If I have a

iddle" class="latex-formula"> chess board, in how many ways can I place four distinct pawns on the board such that each column and row of the board contains no more than one pawn?
Mathematics
1 answer:
SOVA2 [1]3 years ago
7 0

Answer:

576 ways

Step-by-step explanation:

There are 4 choices for the column of pawn in the 1st row

There are 3 choices for the column of pawn in the 2nd row,

There are 2 choices for the column of pawn in the 3rd row, and

There is 1 choice for the column of the pawn in the 4th row

Which gives a total of 4! = 24

Also, the pawns are distinct, so there are 4! ways to place them in these chosen positions;

4! = 24

So, there are 24 * 24 possible ways

= 576 ways

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2 years ago
the utility bill for the Millers home was it in April with $132 42% of the bill was for gas and the rest for electricity how muc
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In the given question, there are several data of immense importance. based on these data's the answer to the question can be easily deduced. it is already given that the total utility bill for the month of April in Millers house is $132 and 42% of the bill was paid for gas. remaining amount is paid as electricity charges.
Now
Amount of money paid for gas = 132 * (42/100) dollars
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Then
The amount of money paid for electricity = (132 - 55.44) dollars
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So the Millers paid 55.44 dollars for gas and 76.56 dollars for electricity in the month of April.
6 0
3 years ago
Read 2 more answers
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