As far as steps i’ll tell you in the comments
Answer:
9.11*10^-4 %
Step-by-step explanation:
To find the probability, you simply need to find the possible outcomes that allows no rooks to be in danger, and the possible amount of ways to place the rooks.
For the first outcome, you start by putting 1 rook in the first columns, you have 8 possible rows to do this. The next rook in the next column will only have 7 possible rows, as you have to exclude the one where the previous rook is located. The next rook, 6 possibilities, the next 5, and so on. So we conclude that the total amount of ways so that none of the rooks can capture any of the other rooks is 8*7*6*5*4*3*2*1 = 8! = 40320
In order to find the total amount of ways to place the rooks, you can just use a combinatoric:
![\left[\begin{array}{ccc}64\\8\end{array}\right]= \frac{64!}{8!(64-8)!} = 4.43*10^9](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D64%5C%5C8%5Cend%7Barray%7D%5Cright%5D%3D%20%5Cfrac%7B64%21%7D%7B8%21%2864-8%29%21%7D%20%3D%204.43%2A10%5E9)
Then:
P = 
Answer:
How many drinks should be sold to get a maximal profit? 468
Sales of the first one = 345 cups
Sales of the second one = 123 cups
Step-by-step explanation:
maximize 1.2F + 0.7S
where:
F = first type of drink
S = second type of drink
constraints:
sugar ⇒ 3F + 10S ≤ 3000
juice ⇒ 9F + 4S ≤ 3600
coffee ⇒ 4F + 5S ≤ 2000
using solver the maximum profit is $500.10
and the optimal solution is 345F + 123S