What is the square number of 4
1 answer:
You might be interested in
Consider ONLY the first class. Since there are n seats (assuming they are in a row), we would have n! ways.
Now, consider the second class. Let's start by arranging three people first and then generalising n people to better understand what is going on.
Where we have 3 people:
First class: A B C = 3!
Second class _ _ _
Now, consider where each of these people can't actually sit.
For A, it's the first seat, for B, it's the second seat, and for C, it's the last seat.
This means that we have a restriction on EACH of the candidates.
So, to tackle this, let's consider A only; B and C will follow the same way.
A can sit in 2 different spots, namely the second and last seat: thereby, having 3C2 ways in sitting. _ A _
Now, when we fix one person, C can only go in one place: first seat. This means that for one single arrangement of the first class, we've made: 3C2 arrangements for the second class, for ONE particular person. Extend that to another person, and we get: 2C1 ways
This extends to what we call: a derangement where the number of permutations made contains no fixed element. We can regard things like picking up three/two/one pen as derangements, because really we're arranging AND not arranging them simultaneously.
Thus, we use the inclusion-exclusion method:
Total no. of perms:
Now, consider the second class. Let's start by arranging three people first and then generalising n people to better understand what is going on.
Where we have 3 people:
First class: A B C = 3!
Second class _ _ _
Now, consider where each of these people can't actually sit.
For A, it's the first seat, for B, it's the second seat, and for C, it's the last seat.
This means that we have a restriction on EACH of the candidates.
So, to tackle this, let's consider A only; B and C will follow the same way.
A can sit in 2 different spots, namely the second and last seat: thereby, having 3C2 ways in sitting. _ A _
Now, when we fix one person, C can only go in one place: first seat. This means that for one single arrangement of the first class, we've made: 3C2 arrangements for the second class, for ONE particular person. Extend that to another person, and we get: 2C1 ways
This extends to what we call: a derangement where the number of permutations made contains no fixed element. We can regard things like picking up three/two/one pen as derangements, because really we're arranging AND not arranging them simultaneously.
Thus, we use the inclusion-exclusion method:
Total no. of perms: