What are you confused about?
Basically, how to do it without choosing numbers. if P can only sit in 1 seat, or two seats, or 3 seats , will have different impacts right? In the solution of this video we have chose numbers, so without choosing numbers i could not do it. For ex: if P can only sit on the two seats, then how do i make sure that that was the correct answer… Actually couldn’t figure out how to do this.
Not really, because those only affect k and n remains invariant (same value).
For instance, if P = 3, then you’d have 3 \cdot 4 \cdot 3 \cdot 2 \cdot 1 arrangements, right?
As another example, for P = 4, you’d have 4 \cdot 4 \cdot 3 \cdot 2 \cdot 1 valid arrangements. It’s hopefully evident that the 4 \cdot 3 \cdot 2 \cdot 1 part is a commonality for all appropriate values of P you choose.
If P=3 that means _ _ _ these tree positions are for P right? so here, if P is in first seat to the left, that means 1*4*3*2*1 is one case: again if P is in the second seat 4*1*3*2*1 and for the third seat: 4*3*1*2*1 … is this how we do it? P can have 3 seats means P can sit in 3 seats right? Whereever he sits among the 3 seats, that is one reserved. Is it?
P = 3 suggests that P has 3 choices of position. Let’s assume, for convenience, that P’s choices involve sitting in the first ,second, or third position.
Case 1: P sits in the first position. Fixing P in this position (P_ _ _ _) gives us 4! ways to arrange the other friends.
Case 2: P sits in the second position. Fixing P in this position (_ P_ _ _) also gives us 4! ways to arrange the other friends.
Case 3: P sits in the third position. Fixing P in this position (_ _ P _ _) again gives us 4! distinct arrangements of the friends.
In essence, we’re just multiplying P’s possible positions (3) with 4!. This makes sense because for each position P occupies, there are 4! arrangements of the other friends.
You can generalize this notion to show that n = 4! = 24