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Alenkasestr [34]
4 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]4 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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How to solve y=2x-15 and y=5x
shepuryov [24]
<h2>Answer:</h2>

By Substitution Method

<h2>Step-by-step explanation:</h2>

Solve by Substitution :

We have two equations

    y = 5x     ---------------> Equation 1

    y = 2x - 15  ------------> Equation 2


   y= 2x - 15  and y= 5x

So Put the value of y in Equ 2

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Solution :

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4 years ago
For a fundraiser, Marcus needs to raise $600 in order to meet his goal. Marcus will walk a certain number of miles, and he will
aliya0001 [1]

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Step-by-step explanation:

Step 1:

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Step 2:

Given, x denotes the number of miles walked by Marcus for the fund raiser

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Step-by-step explanation:

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Answer:

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