If the 4 states have to be in a specific order say ABCD,
then the total number of different possible routes is:
43P4 = 2,961,840
So the probability is:
1 / 2,961,840 = 3.38 x 10^-7
But if the 4 states can be in any order such as DBAC,
ACBD etc, then the total number of different possible routes is:
43C4 = 123,410
So the probability is:
1 / 123,410 = 8.1 x 10^-6
No I don’t think it is practical to list all the
different possible routes to select the one that is best. We can simply use
mathematical models to solve for that one.
<span> </span>
I think it will take the ball
Answer: There are 17576000 ways to generate different license plates.
Step-by-step explanation:
Since we have given that
Numbers are given = 0 to 9 = 10 numbers
Number of letters = 26
We need to generate the license plate numbers.
Since there are repetition allowed.
We would use "Fundamental theorem of counting".
So, the number of different license numbers may be created as given as

Hence, there are 17576000 ways to generate different license plates.
The required matrix is:
![\left[\begin{array}{ccc}-25&17&0\\8&-1&3\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-25%2617%260%5C%5C8%26-1%263%5C%5C%5Cend%7Barray%7D%5Cright%5D)
Step-by-step explanation:
We need to apply elementary row operation -2R₂+3R₁ tothe matrix:
![A=\left[\begin{array}{ccc}-3&5&2\\8&-1&3\\\end{array}\right]](https://tex.z-dn.net/?f=A%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-3%265%262%5C%5C8%26-1%263%5C%5C%5Cend%7Barray%7D%5Cright%5D)
Multiplying Row 2 with -2 and Row1 with 3 and adding,
-9 15 6
-16 2 -6
----------
-25 17 0
After applying this operation, Row 1 will be changed while Row 2 will remain same because we get -2R₂+3R₁ -> R₁
The required matrix is:
![\left[\begin{array}{ccc}-25&17&0\\8&-1&3\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-25%2617%260%5C%5C8%26-1%263%5C%5C%5Cend%7Barray%7D%5Cright%5D)
Keywords: Matrices, elementary row operation
Learn more about matrices at:
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