In how many ways can 5 girls (A, B, C, D, E) can sit on five chairs (1, 2, 3, 4, 5) if girl A does not want to sit on chair number 1 or chair number 2 and girl B wants to sit on 3rd or 4th chair?
This question was dealt in a lecture which says I should satisfy a "positive condition" first only then I should look at the "negative condition". The positive and negative conditions seem little blurry mathematics to me.
Here is how it was dealt with:
B can sit on 2 chairs (positive condition)
A can sit on 2 remaining chairs (negative)
and the rest of them can sit in 3! ways
So $2\cdot2\cdot3\cdot2\cdot1$ ways
But when I consider seats instead of people as reference
There are 3 ways in which chair number 1 can be filled ($\{C,D,E\}$)
There are 2 ways in which chair number 2 can be filled ($\left|\{C,D,E\}\right|-1$)
There are 3 ways in which chair number 3 can be filled ($\left|\{A,B,C,D,E\}\right|-2$)
There are 2 ways in which chair number 4 can be filled ($\left|\{A,B,C,D,E\}\right|-3$)
There are 1 ways in which chair number 5 can be filled (remaining girl)
So, $3\cdot2\cdot3\cdot2\cdot1$ (for seat number one to five respectively)
I have two questions regarding this problem. Is there some better way to think of the approach given in lectures other than positive/negative? Does this concept has any name?
Another question is why doesn't the latter method (the one considering seats instead of people) work?
Best Answer
In your way of counting, if A sits in chair 3 you must have B in chair 4 so the choices are not independent. I don't see the advantage in the distinction between positive and negative conditions. The same problem could be posed that A wants to sit in $3,4, \text { or }5$ and B will not sit in $1,2, \text { or } 5$, which interchanges the positive and negative conditions. The key point is that whichever chair you select for B, there are the same number of options for A. If you seat A first, the number of options for B is not constant. You can divide it by cases, but seating B first avoids that.