Given six people of different heights, how many ways can we order them so no 3 consecutive people are ordered in increasing height

141 Views Asked by At

https://artofproblemsolving.com/wiki/index.php/Principle_of_Inclusion-Exclusion#Four_Set_Example Here is where I found the problem, go to the four set example section. My question how do they say |A| is 120. I am very new to combinatorics and counting, so I apologize if I my fundamental understanding is just flawed. A is the event where People 1,2,3 are in ordered height, meaning that People 4-6 can be moved around. Since We can only move around 4-6 wouldn't |A| just be 3!? They say that it is (6 3), how many combinations of 3 items from 6 items. I'm assuming that 3! is for people 4 - 6 and the combination (6 3) for People 1-3. Hope my question made sense.

1

There are 1 best solutions below

2
On

Let us assume that the people in increasing order of height are p1,p2,p3,p4,p5 and p6.

In scenario A, the people standing in first 3 places are in increasing order of height. But they need not necessarily be p1, p2 and p3. so, first we need to pick three people to arrange in increasing height, which we do using (6 3). This order may be {p1,p3,p5} or {p2,p3,p4} we just need to choose any three random people and arrange them in only one possible way which is in increasing order of height. Then we arrange the remaining three people standing in the fourth, fifth and sixth place in 3! ways.