Question

Permutations questions? Four in total..

This may be confusing, but I am going to try and write it out.. The "E"s are supposed to represent Greek E's.

5
E 2n+1 evaluate
n=1

3
E 3n evaluate
n=1

Type the formula for the number of permutations of n things taken r at a time.
P(n,r)=?

Type the formula for the number of circular permutations of n things.
Pc(n,n) = ?

Thanks y'all for explaining!

Answer 1

Let's take this a little bit at a time. First of all, how to do summations.
\[\Large \sum_{n=1}^{5}2n+1\]
means "Summation from n=1 to n=5 of (2n+1)"

Answer 2

What you do is you start at n=1 and you go up by 1 each time until you get to 5. So, n=1, then n=2, then n=3, then n=4, and finally n=5.
At each n, you do the 2n+1, and you add it to the total.

Answer 3

Any questions? Try it and let me see your attempt.

Answer 4

Oh crud. Okay. Hmm.
You have the first summation. Starting with 1, you substiute it in. 2(1)+1=3, 2(2)+1=5, 2(3)+1=7, 2(4)+1=9, 2(5)+1=11. Then you add them all? 3+5+7+9+11=35. So your answer is 35? Am I close..?

Answer 5

So goooood! You're a boss.

Answer 6

Try the next summation.
\[\Large \sum_{n=1}^{3}3n\]

Answer 7

Hahahaha. I've been told. Just kiddingg. ;)
And okay. Let's go..
You will go from 1-3 because 3 is at the top. Subsituting starting with 1 is.. 3(1)=3, 3(2)=6, 3(3)=9. 3+6+9=18. Maybe? What does the n=1 at the bottom mean? What I have noticed is that is almost always 1.

Answer 8

It does two things. It tells you which variable is changing each time, so in this case it's n. Sometimes it might say x=1 or i=1, which would mean that x or i was changing. The other thing is that it tells you which number to START at, so if it doesn't say n=1, but n=4, then you would start at 4 instead of 1.

Answer 9

Ohh. Okay. That's probably a good thing to know :) Thanks!

Answer 10

And yes, most examples will start at 1, but especially later on, they won't all start at 1. Let me show you an example to help you see one that ISN'T like that:
\[\Large \sum_{i=3}^{6} 2i = 2(3) + 2(4) + 2(5) + 2(6)\]
= 6 + 8 + 10 + 12 = 36

Answer 11

Holy moly. Well, I'm glad I'm not at that yet.. Haha

Answer 12

And thank you soo much!! :) I really appreciate it!

Answer 13

Okay, next part. Permutations.

Answer 14

Oh great. But no time like the present I suppose.

Answer 15

I'd like to use an example. Imagine that I have a bag with 5 marbles in it, and they are 5 different colors.
How many possible colors could I pull for the first marble?

Answer 16

5 colors

Answer 17

Cool. So 5 possibilities if I'm just drawing one marble. Now, imagine I've pulled one marble. How many possibilities are there for the second marble I draw?

Answer 18

4 colors

Answer 19

Exactly. So if I pull 2 marbles, there are 5 possibilities for the first marble and only 4 for the second marble.
So the number of possible outcomes are
5*4

Answer 20

There are 20 possible outcomes if I pull 2 marbles.
red blue
red green
red orange
red yellow
blue red
blue green
and so on.

Answer 21

Okay. And if you kept going it would be 3 then 2 then 1? So It would be like 5! right?

Answer 22

Ah, yes! You've studied factorials =)

Answer 23

So think of it this way.
If I pull all 5, then it's 5! which is 5*4*3*2*1.
But what happens if I don't pull all 5, but only 3 marbles? Well, then I don't want to do 5*4*3*2*1, I want to do just 5*4*3.

Answer 24

Or if there were 10 marbles and I pulled 3, then it would be 10*9*8

Answer 25

Yes I have studied them, I'm in the middle now actually. And if you didn't pull them all you would only multiply the first x number? x=number of marbles pulled. Right?

Answer 26

Very good =) Exactly. Now, what I'm going to show you next is just a fancy way to write that as an equation, but you have the basic concept.

Answer 27

Let's say I have 10 marbles and I pull 4.
If I do 10!, that's too much. I need to stop before multiplying all of those. There's a simply way to write an equation that says WHEN to stop.

\[\frac{10!}{6!} = \frac{10*9*8*7*6*5*4*3*2*1}{6*5*4*3*2*1}\]

Answer 28

Okay, so I know that last step maybe looks just random and crazy, but look what happens when I divide 10! by 6!. Do you see how 6 is in the top and bottom? Do you see how everything after 6 is also in the top and bottom?

Answer 29

Yes. And so it cancels out leaving only what you need. 10*9*8*7, right?

Answer 30

Exactly right =)
So when I wrote
\[\frac{10!}{6!}\]
it was a fancy, but neat way of writing 10 factorial, but stop at 7, don't do 6 or anything after.

Answer 31

Now, let's try to answer their question
Type the formula for the number of permutations of n things taken r at a time.
P(n,r)=?
What if there are n marbles in the bag, and we pull r of them?

Answer 32

p=n!/r!?

Answer 33

Would that work for our 10 marble bag when we pulled out 4? It would be 10!/4!, which would mean we would do 10*9*8*7*6*5

Answer 34

Hmm. No, so.. P=n!/(n-r)! maybe..?

Answer 35

That's the only way I could think to get 6 on the bottom.

Answer 36

Oh my goodness. So impressive.

Answer 37

Wait. Really? No way.

Answer 38

Really. Has anyone ever told you that you're brilliant at math?

Answer 39

No. There is probably a reason for that. Haha. And thank youuu!!!