Question 285534: Provide an example of a situation that describes a permutation?
Provide a situation that describes a combination?
How many different plates are possible in the United Kingdom?
Identify the probability that a randomly selected plate is AA 51 ZZZ?
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! 4 runners in a race.
Only 3 will get a prize.
What are the possible ways the prize can be distributed?
Since order is important, this will be a permutation.
You will determine how many ordered sets of 3 can you draw from the 5 runners.
The formula will be:
n! / (n-x)! where n is the total number of runners and x is the number of runners in the set that you will be drawing.
n = 4
x = 3
n! / (n-x)! = (4*3*2*1) / (1) = 4*3*2 = 24
If we give each runner a letter, then the possible letters for the 4 runners are:
a,b,c,d
The possible sets of 3 that we can draw are:
a,b,c *****
a,c,b
b,a,c
b,c,a
c,a,b
c,b,a
a,b,d *****
a,d,b
b,a,d
b,d,a
d,a,b
d,b,a
a,c,d *****
a,d,c
c,a,d
c,d,a
d,a,c
d,c,a
b,c,d *****
b,d,c
c,b,d
c,d,b
d,b,c
d,c,b
What are the possible ways in which runners will be eligible for a prize?
Since order is not important, this will be a combination.
You will determine how many unordered sets of 3 can you draw from the 5 runners.
The formula will be:
n! / ((n-x)!*x!) where n is the total number of runners and x is the number of runners in the set that you will be drawing.
n = 4
x = 3
n! / ((n-x)!*x!) = 4! / (1!*3!) = (4*3*2*1) / (1*3*2*1) = 4
The 4 possible ways are:
a,b,c
a,b,d
a,c,d
b,c,d
Notice that the 4 possible ways the runners will be eligible for a prize is the asterisked permutations above.
In other words, if you disregard the order in which they came in, one of the possible combinations is a,b,c.
If you then wanted to determine the possible ways they could come in, you would then get:
a,b,c
a,c,b
b,a,c
b,c,a
c,a,b
c,b,a
The same 3 runners could come in 6 different ways.
The combination got you who the runners were.
The permutation got you who the runners were and the order in which those same 3 runners could come in.
If you will notice in the formulas:
The combination formula was:
n! / ((n-x)! * x!)
The permutation formula was:
n! / (n-x)!
The x! in the divisor of the combination formula is missing in the permutation formula.
That x! is the number of ways the same set of runners could finish.
There were 3 runners and they could finish in 3! ways.
3! = 3*2*1 = 6 ways the 3 runners could finish.
The number of combinations was 4! / (1!*3!) = 24 / 6 = 4.
The number of permutations was 4! / 1! = 24 / 1 = 24.
Combinations only got the people regardless of how they finished.
Permutations got the people and how they finished.
LICENSE PLATES IN GREAT BRITAIN
The possible letters do not add up to the total letters in the alphabet but I don't think that's what you want.
See this reference for what I am talking about:
http://en.wikipedia.org/wiki/Vehicle_registration_plates_of_the_United_Kingdom#Current_system
Assume that all letters are valid, each character can have 26 letters and each number can have 10 numbers.
The total number of possibilities would be:
26*26*10*10*26*26*26 = 26^5*10^2 = 11881376700 total possible license plate numbers.
The probability of getting AA 51 ZZZ would be 1 out of 11881376700 = 8.416533573 * 10^-10
Looked at another way, the probability would be:
1/26 * 1/26 * 1/10 * 1/10 * 1/26 * 1/26 * 1/26 = 1/26*26*10*10*26*26*26) = 1/(26^5*10^2) = 1/1188137600 = 8.416533573^(-10).
|
|
|
