Permutations Flashcards
What is the result of the following factorials?
0!
1!
2!
3!
4!
5!
6!
0! = 1
1! = 1
2! = 2
3! = 6
4! = 24
5! = 120
6! = 720
In Permutations, which element is the lines we use?
The shortest of the more scarce element is always the number of lines
What happens if N and K are equal?
Meaning the number of lines and the subject?
N = K it means N!
What is the formula for when there are circular arrangements?
(N-1)!
How do you solve permutations with repeating elements?
Example: In how many ways can the letters Mississippi be arranged in a row?
Chat GPT A2 Problems
When there are repeating elements, you need to put those repeating elements in the denominator.
Numerator = 11!
Denominator = 1! (For M), 4! (For i), 4! (For S), 2! (for P)
11!/ 1!4!4!2!
= 34,650
Part 1
How do we solve for permutations where K is changing?
First, How is the table we have to learn composed?
For addition look for the following:
- Or
- At least
- At most
-Complete results
For multiplication look for the following:
- And
- Default
- Partial Results
Solve two Chat GPT problem - Permutations with K is changing
Permutations with restrictions.
Which are the four common restrictions?
Chat GPT 4 problems - Permutations with Restrictions
1.- Circular arrangements = (N -1)!
Example: In how many ways can 5 people be seated at a circular table?
(5-1)! = 4!
2.- Couples or groups that must sit together = One Lines for each Group, and then Sub lines for each group
Example: Group |ABC| |DE| |FGH| How many possible ways can they sit?
3! = For each group (there are 3 groups)
Group 1 |ABC| = 3!
Group 2 |DE| = 2!
Group 3 |FGH| = 3!
3!3!3!2! = Result
3.- Alternating Seating = Depends if its Even or ODD
Odd Example: How many different ways can 3 boys and 2 girls can be seated if they can to be alternated?
In this case, the boys have to start and end the seating for the alternation to be completed.
_3 _2 _2_1_1= 3!2!
Even Example: How many different ways can 3 boys and 3 girls can be seated if they can to be alternated?
In this case, either boys or girl could start
_6 _3 _2 _2 _1 _1 = 3!3!2 = 72
4.- Limited Seating = John Must Sit Between two friends
Example: John and 4 friends go to a lakers game. In how many ways can they be seated in 5 consecutive seats, if John has to sit between 2 friends?
We will calculate the invalid scenarios because they are easier:
There are two invalid scenarios, each in which John sits at the beginning or end of the seats. We will use J to identify John
Invalid Case 1:
_J _4 _3 _2 _1 = 4! = 24
Invalid Case 2:
_4 _3 _2 _1 _J = 4! = 24
All Possibilities:
_5 _4 _3 _2 _1 = 5! = 120
All Possibilities - Sum of Invalid Cases
120 - 48 = 72 ways John could sit
What is the best and Fastest way to respond to permutations with restriction?
Let’s use this example:
John and 4 friends go to a Lakers Game. In how many ways can they be seated in 5 consecutive seats, if John has to sit between 2 friends?
A) 24
B) 72
C) 120
D) 240
E) 720
You know 5! = 120 and It has to be less than that, so we eliminate C,D,E
A is too small so B has to be the answer
What is the Decision Tree for permutations?
1.- Are items allowed to be repeated? License Plate
2.- Are any of the elements identical? Mississippi
3.- Is K changing in the problem? Anna
4.- Are there any restrictions? Fantastic 4
