Permutation/combination Destroyer Test 4

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emminent

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Can you guys explain your way of doing permutation/combination that you found easiest for you to understand.

Having trouble understanding these two questions even though I read the explanation.

A theater has 5 doors. in how many ways can an actor enter one door and leave in a different door?

A piece of paper has 10 distinct dots drawn on it. if a pair of dots determine a line, how many different ways are there to draw a line on this paper?
 
The way I remember it is for problems where order does matter, it's simple. For problems where order doesn't matter it's a bit more complicated.

For the theatre question, you can see that order matters because entering through door 1 and leaving through door 2 is NOT the same as entering through door 2 and leaving through door 1. Therefore, this is one of the "simple" questions. You have 2 events (2 doors being utilized), so your equation will be
__ x __
Now you just have to figure out what two numbers to plug in. There are 5 doors total, so our first choice would have 5 possibilities, and our next choice would then have 4 possibilities (because you can't exit through the same door you came in.)
5 x 4
This is one of those questions where as long as you realize that order matters, you can write it out every possible combination
(1,2 1,3 1,4 1,5
2,1 2,3 2,4 2,5
3,1 3,2 3,4 3,5
4,1 4,2 4,3 4,5
5,1 5,2 5,3 5,4) and still get the correct answer.


For the dot question, you can see that order doesn't matter because connecting point 1 to point 2 will get you the same exact line as connecting point 2 to point 1, so this will be one of the complicated questions. I approach this problem a different way than most sources just because the equation confuses the crap out of me, but if my explanation doesn't make sense to you then just try understanding the combination equation.
Just like the problem before, you have 2 events (2 dots being chosen), but now you have 10 possibilities (10 different dots).
__ x __
For the first number you have 10 possibilities, for the second number you have one less.
10 x 9
BUT since order doesn't matter here, this number will be higher than the correct answer. What we've calculated so far includes a line from point 1 to point 2, and another line from point 2 to point 1, which we've already realized will be the same line. Therefore, we've overshot. (This is where my way of solving the problem strays from the norm.) We need to divide our current answer by how much we've overshot. The way I look at is that we have 2 events, so divide by 2 factorial.
(10 x 9)/(2!) = (10 x 9)/(2 x 1) = 45
This isn't the best problem to use as an example to understand this concept so let's pretend the problem said you had to connect 3 points to make a line, instead. In that case, we would divide our final answer by 3 factorial because dots 1, 2, 3 would form the same lines as
dots 1, 3, 2
dots 2, 1, 3
dots 2, 3, 1
dots 3, 1, 2 and
dots 3, 2, 1
That means we've actually counted the same line 6 different times. Then our answer would be (10 x 9 x 8)/(3!) = (10 x 9 x 8)/(3 x 2 x 1) which is the same thing as (10 x 9 x 8)/(6).


I recommend doing a bunch of practice problems until you get the hang of when order counts and when it doesn't counts, and what you should do in each situation. Most of the time, as long as you understand what they're asking for you can write out each possibility and count them up if you forget the equation. Hope that helps!
 
Thanks danger.

How about this question.

How many chords on a circle can be drawn from 6 distinct points lying on the circle?

So there's 6 possibilities which is 6! and then since there's two events we divide by 2!. But where does the 4! come into play? I don't get that part.
 
Alright I see where you are confused. The textbook way of doing this problem is
n=6 and r=4 so your answer is (6!)/(4! x 2!)
If you evaluate that, you get (6 x 5 x 4 x 3 x 2 x 1)/(4 x 3 x 2 x 1 x 2 x 1) which ultimately cancels out to (6 x 5)/(2 x 1) = 15

This approach never really made sense to me, so I use a modified method. It's important to remember that for this method, the numerator is NOT a factorial.
A chord consists of two points, so our numerator is
__ x __
We have 6 possibilities for the first point, and then 5 possibilities for the second point
6 x 5
Since order doesn't matter, we've over-counted. Divide your current answer by the number of events factorial.
(6 x 5)/(2!) = (6 x 5)/(2 x 1) = 15
If you compare this method to the textbook one, we've skipped the step having to deal with the confusing 4! part because that part will cancel out anyway.
 
Thank you Heisenberg. I wasn't able to apply myself initially but I think I'm getting the hang of this
 
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