In chess, a rook attacks all cells on the row and column that it is in. let k ≤ n. how many ways are there to place k rooks on an n × n chess board in such a way that no two rooks attack one another?

Step-by-step explanation:

First, we select the columns from our n columns where we will place our k rooks. We use a combinatoric for this:

Using the same line of reasoning, we choose the rows to place our k rooks from our n rows:

Now, we need to form cells using the colums and rows we selected. For example, imagine a 3x3 board, where you will place 2 rooks. We select the first 2 columns and the firsts 2 rows to place the rooks. The cells where the rooks are gonna be placed are formed by pairing columns and rows, like this:

- First choose the Column 1. You have 2 rows available for pairing: Row 1 and Row.

- Once you formed the first cell, you will have left 1 column and 1 row left to place the remaining Rook.

Extrapolating this logic for k Rooks, you will notice that the number of possible ways to distribute the rooks will always be k!.

Finally, the total amount of ways there are to place k rooks on an n × n chess board in such a way that no two rooks attack one another will be the multiplication of the 3 expressions we develop so far:

Total = * * k! =

%40

step-by-step explanation:

d. 24+-0.114

step-by-step explanation:

the question requires us to calculate the approximate 95% confidence interval for the mean number of ounces of ketchup per bottle in the sample given the following statistics;

sample mean = 24

standard deviation, σ = 0.4

sample size , n = 49

the confidence interval for a population mean is calculated using the formula;

sample mean ± z-score*(σ/sqrt(n))

the z-score associated with a 95% confidence interval, from the standard normal tables, is;

1.96

substituting these values into the formula;

24 ± 1.96*(0.4/7)

= 24 ± 0.112

from the alternatives given, choice d is closest to the above expression;