Question

Help with Permuations and Combinations.

Answer 1

7. 4!/2!


9. 7Cv2


11. 6vPv2

Answer 2

@triciaal @TheSmartOne

Answer 3

not now, sorry

Answer 4

First one is a factorial.. do you know how to deal with factorials?

Answer 5

No. . .

Answer 6

Factorials are just products indicated by an exclamation mark.

For example, \(\large 6!=6*5*4*3*2*1 \)

Answer 7

Oh yeah, I know how to do that.

Answer 8

I just don't know how to do them when they aren't one digit. xD

Answer 9

Try breaking the factorial apart to match something so you can cancel them. The 4!=4*3*2*1. The 2!=2*1.

You can stop at: \(\Large \frac{4*3*2!}{2!} \)

Now you can cancel the 2!

Answer 10

So. . .4 * 3

Answer 11

Yes, and 4*3=12, and that's pretty much all there is to it :)

Answer 12

Oh. . .

Answer 13

Thank you! This is what I have so far for #9.

7!/2!(7-2)! =7!/2! *2!

Answer 14

How did you get another 2!? Isn't 7-2=5? :P

Answer 15

I don't know, honestly. xD
So. . .

7!/2!(7-2)! =7!/2! *5! =17

Answer 16

Wait. . .

Answer 17

Nevermind, I've confused myself. xD

Answer 18

Try breaking the 7! to match the biggest factorial in the denominator

Answer 19

How do I do that?

Answer 20

Well you know how 7! is just 7*6*5*4*3*2*1? Trying taking out numbers. Like if you wanted to take out 7 from it, take it and be left with \(7*(6*5* \) 4*3*2*1).. everything else can just represented as a factorial: \(7*6! \)

So for yours we have \(\Large \frac{7!}{2!*5! }\)

Let's break up the 7! to match 5! sooo \(7*6*(5*4*3*2*1) \)

Or just : \(\Large \frac{\color{green}{7*6*5!}}{2!*5! }\) Now we can cancel

Answer 21

7*6!/ 2!. . .

Answer 22

Actually, the 5 is the only one with the factorial, the others are just numbers, so it's (7*6)/2!

Answer 23

(7*6)/2!. . .Then what?

Answer 24

Since 2! is just 2*1, and that's 2 either way, just divide \(\Large \frac{7*6}{2} \) :)

Answer 25

=21 : D

Answer 26

Yes :)

Answer 27

6*5*4*3*2/2!

Answer 28

. . .Not sure if I'm doing it right

Answer 29

It's the third one right?

Answer 30

Yes.

Answer 31

\(\Large nP_{k}\frac{n!}{(n-k)!}\)

\(\Large 6P2 =\frac{6!}{(6-2)! }\)

:P

Answer 32

Of course, that's why we have formulas. x3

Answer 33

So you were right with the process, just got the wrong number :P

Answer 34

Thanks for your patience, you helped a lot.

Answer 35

You're welcome :) \(\Huge \color{\green}{\star} \)

Answer 36

Good job @Luigi0210 (: