Question

You don't need to show me how to do all 4 problems unless you're willing but I'd appreciate it if you at least informed me how to solve the concept of the problems

http://awesomescreenshot.com/0d52et4553

Answer 1

It looks like you're dealing with combinations and permutations.
The formula for a permutation (P) is: \[_nP_r= \frac{n!}{(n-r)!}\]
The formula for a combination (C) is:\[_nC_r=\frac{n!}{r!(n-r)!}\]
I don't have time to work each one out, but you should be able to plug them into the formulas.

Answer 2

I forgot what the exclamation marks are indicating me to do...

Answer 3

exclamation marks mean factorial. \[n! = n(n-1)(n-2)(n-3)(n-4)...(2)(1)\] You basically multiply by all whole numbers equal and less to that number. So, 5! is 5 x 4 x 3 x 2 x 1, or 120.

Answer 4

\[_{6}P_{3} = \frac{ 720 }{ (6-3)! }\]
Sorry to be bothering you but do I have to subtract the values in the denominator before factoring? Or do I factor both numbers and then subtract?

Answer 5

By chance do any of you 3 know what to do lol

Answer 6

Which three are you referring to? ^_^

And of course you have to subtract first before applying factorial, that's what those parentheses are there for, after all :)

Answer 7

You and the two other viewers although it's actually you and another now lol... I swear I'm not crazy /.\
and thank you... so....
\[\frac{ 720 }{ 6 } = 120\]

Answer 8

I'm iffy with my symbols btw lol

Answer 9

Well, there's actually no reason to be so dogmatic with formulas.
Maths is there for you to provide shortcuts for yourself (creatively and correctly, of course)

A quicker way (in my opinion) to evaluate 6P3 is to just multiply from 6 downwards UNTIL you have exactly *3* factors.
6 x 5 x 4 = 120
Simpler, without all that ! stuff :3

Answer 10

This works with any such nPr, by the way :3

Answer 11

Try 9P4 using this formula first... (9 factorial? Unbelievably large :3)....that's if you can...

And then try using this shortcut... multiply from 9 downwards UNTIL you have exactly *4* factors.

Answer 12

9x8x7x6= 3024? o. o

Answer 13

XD everything related to math makes me URGH unless I get it right on the first time... haven't done so in years unfortunately...
But...
\[\frac{ 2177280 }{ 90720 }\]

I feel like I messed up somewhere terribly e.e

Answer 14

Oops... I meant
\[\Large \frac{9!}{(9-4)!}\]
My bad :)

Answer 15

By the way, just google 9! and stuff...

Answer 16

xD it's ohtay... ummm

\[\frac{ 2177280 }{ 3024 } = 720\]

o.o *goes to google it*

Answer 17

LOL it's wrong... can you now see just how atrocious it is to try to evaluate large factorials? XD

Answer 18

9! is in fact 362880
now (9-4)! is just 5! or 120

Answer 19

O___O

Answer 20

So, try dividing 362880 by 120
:)

Answer 21

omgeez /.\ ooooh I put it in the calculator wrong, huh?

18144

Did I put the numbers in correctly this time? >-< lol

Answer 22

Probably.
Try again.
9! divided by 5!
Or 362880 divided by 120

You really have to get the correct answer... unless you're keying it in incorrectly.

Answer 23

362880 / 120 = 3024

o.o Sometimes when i divide I tend to plug in the numerator and the denominator in the incorrect places >.<

Answer 24

omg please tell meeeeee it's righhht

Answer 25

I don't HAVE to tell you it's right. You should KNOW it's right, since you got the same answer originally by just doing it the way I do factorials...
9x8x7x6

remember? You got 3024 then, didn't you? ^_^

Answer 26

DANG IT LOL I can't with this -flips table-
You are correct my good...sir/ma'am... lol -facepalm-

Answer 27

We're not done yet, we still have the nCr s to worry about.
I'm a boy, and for the love of humanity, don't call me 'sir'.

It's *TJ* ;)

Answer 28

XD Lo siento TJ ... How do you do?
And oooh...goodie... /.\ more trauma

Answer 29

alrighty

Answer 30

Crud, I made that typo again -_-
\[\Large _9P_4 = \frac{9!}{(9-\color{red}4)!}=\frac{9!}{5!}\]
Okay?

Answer 31

XD it's all G... ok o.o

Answer 32

Now, let's expand.
\[\Large =\frac{9\cdot8\cdot7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1}{5\cdot4\cdot3\cdot2\cdot1}\]
And actually, a huge chunk of the numerator gets cancelled out...\[\Large =\frac{9\cdot8\cdot7\cdot6\cdot\color{blue}{5\cdot4\cdot3\cdot2\cdot1}}{\color{blue}{5\cdot4\cdot3\cdot2\cdot1}}\]

\[\Large =\frac{9\cdot8\cdot7\cdot6\cdot\cancel5\cdot\cancel4\cdot\cancel3\cdot\cancel2\cdot\cancel1}{\cancel5\cdot\cancel4\cdot\cancel3\cdot\cancel2\cdot\cancel1}\]
Effectively leaving...
\[\Large 9\cdot8\cdot7\cdot6\]
As it should be :P

Answer 33

Oooooh yeaaa I understand o.o

Answer 34

Now, to nCr
Do you know the formula?

Answer 35

Well, even if you do know it, I'll post it anyway :)
\[\Large _nC_r = \frac{n!}{\color{blue}{r!}\cdot (n-r)!}\]
I coloured in blue the difference it has with the formula for nPr

Does this look daunting? :)

Answer 36

Oh wait, it was already posted... derp :3

Answer 37

Not by heart but I can just scroll up lol... it does actually. It really does /.\

Answer 38

Orrr you can just type it for me xD

Answer 39

Shhhshhshh it's G

Answer 40

ok it looks just a little less complex... just a bit lol

Answer 41

Okay... this isn't intuitive... let's call this n CHOOSE r :)

There's a reason for this.
What it actually does is count how many ways you can *choose* r items from a set of n different items.

Say, you need two lab partners from a set of 5 students.
Then the answer is simply 5C2.

Feel any better? :3

Answer 42

... I suppose... except I know how great a number this could turn into with bigger outcomes! OMG!

Answer 43

That's assuming the 5 students are different of course, hopefully that's a given (the prospect of 5 completely identical students is rather unnerving O.O )

Now then... to business...

Answer 44

LOL that's... something to think about [x

Answer 45

Okok... *serious face*

Answer 46

Okay, enough fooling around from me, let's do this...

Answer 47

Let's solve the third item using the formula first, shall we?

Answer 48

5C2.
\[\Large _5C_2 = \frac{5!}{2! \cdot (5-2)!}\]
Please solve ^_^

Answer 49

I'll gladly ATTEMPT to solve lol...
\[_{5}C_{2} = \frac{ 20 }{ 6 }\] >.< oh goodness

Answer 50

wait wait

Answer 51

Nope.... you should ALWAYS get an integer solution for such problems... because there are counting problems... silly Angel :P

Answer 52

OH MY GOODNESS these rules will be the death of me ;-;
May I ask...
What exactly's an integer solution?
Sorry for all of my questions and struggles >.\

Answer 53

A whole number?
Which 20/6 is most certainly NOT.

Try again. ^_^
I know you can do this :)

Answer 54

Urghhhh okay... lemme see...

Answer 55

So...
In the numerator and the denominator the factors can be cancelled up to 3 which leaves me with
\[\frac{ (5)(4) }{ 2! } \]

because the 3 was cancelled out... except... the two factored is just 2 and 20/2 = 10 but it doesn't look right to me... did I mess up somewhere >.<

Answer 56

10 is correct.
That was... exhausting :D

Answer 57

So... by now, you're probably expecting a sort of shortcut from me or something, much like with nPr... are you? :3

Answer 58

Eh... mathematic calculations are so subtle when there's no confidence in understanding the concept of the problem *sigh*... thank you lol... sorries.
>.>
Are you by chance insinuating something?

Answer 59

HmHm?? ;b

Answer 60

Of course. There is a faster way of doing this.
Much like with nPr, multiply from 5 downwards UNTIL you have exactly 2 factors.

The only difference is, you divide that answer with 1 multiplying UPWARDS until you have exactly two factors.

5x4
divided by
1x2

20/2
10
Voila :P

Answer 61

>.< wait whaaaat

Answer 62

Where did the 4 come from o.o

Answer 63

5 multiplying downward?
What number comes directly before 5? -_-

;)

Answer 64

4 o.o lmaoooo but...

Answer 65

wait so if it was 7C3

it would be 7 x 6 ?

Answer 66

No... because you multiply downwards until you have *3* factors then :P

Answer 67

DANGNABBIT

Answer 68

May I have one more example?

Answer 69

Of course.
Let's finish up 7C3 first, though.
That would be
7x6x5
divided by
1x2x3

If I'm not mistaken, that's 35 :P

Answer 70

Thank you and mistaken you most certainly are not... T:

Answer 71

Now, let's have a look at a more complicated case...

Try 9C8 :P

Answer 72

This can't be good... by my method, it would seem like you have to multiply from 9 downwards until you have eight factors...
right?
9x8x7x6x5x4x3x2

Simply nasty, isn't it? ;)

Answer 73

>-> woah dere

Answer 74

>.< throw a potato at me why don't you lol

Answer 75

Well, do you not trust me to make this easy for you? :D
Believe it or not, 9C8 is equal to 9C1

Do you see why?

Answer 76

Trust issues LOL
ummm...
does it have anything to do with the difference between 9 and 8 being 1

Answer 77

WAIT

Answer 78

Yes it does :P
Think about it, sans all that maths complications.
Remember...

Answer 79

9C8 is how many ways you can CHOOSE 8 items from a set of 9 different items, yes?

Answer 80

I'm confused with all of this item talk lol /.\
When you're doing your little tactic over there lol are you saying that I factor both n and r by the number that's r?

Answer 81

Wait, just bear with me... instead of items, why don't we just say colours? :P

If you have 9 colours and you are meant to choose 8, then... does that sound better?

I'm trying to make you *feel* them essence here...also, I'm bored :3

Answer 82

Ooooh makes sense and lol glad you're bored otherwise I would have been left to fend for myself ;-; I'm a handful smh

-puts hands up- I FEEL IT

Answer 83

Now, if you are going to choose 8 colours from a set of 9 different colours, it's as if you're just choosing WHICH ONE you're NOT going to choose, right?

Answer 84

o.o
I suppose...

Answer 85

That's why, the number of ways to choose 8 colours from a set of 9 is exactly the same as the number of ways to choose *one* colour from a set of 9.

Answer 86

Alternatively, say we need to choose 7 colours from a set of 9.
Then that'd be the same as choosing which *TWO* colours you'd be leaving out.

Hence, the number of ways to choose 7 colours from a set of 9 is exactly the same as the number of ways to choose *two* colours from a set of 9.

Answer 87

Lost? :)

Answer 88

Try to understand it, though if you simply can't, then just tell me...

Answer 89

And we'll just move on ^_^

Answer 90

7 colors from a set of 9 ---> 9 x 7 / 7 x 6
Shoot idk...
We can move on /.\

Answer 91

Nope... Anyway, maybe we should leave that (ehem) beautiful conceptual stuff for when you 've mastered the formulas.

Right now, just accept that
\[\Large _nC_r = _nC_{n-r}\]

Answer 92

\[\Large _nC_r =\frac{n!}{r!\cdot (n-r)!}\]\[\Large _nC_{n-r} =\frac{n!}{(n-r)!\cdot [n-(n-r)]!}=\frac{n!}{(n-r)!\cdot n!}\]
Clearly equal. :P

Answer 93

Sorry, I'm just typo-prone right now :/

Answer 94

So, with that in mind, back to the problem
9C8

We now know this to be equal to
9C(9-8) = 9C1
right? ^_^

Answer 95

No apologies necessary lol
and
yesh

Answer 96

So, multiply downwards from 9, until you have exactly 1 factor....
so trivial...
just 9, right?

divide this by 1, multiplied UPWARDS until you have exactly 1 factor... so... just 1, then :P

9/1
= 9

Problem solved.

Now try doing it by formula O.O

\[\Large _9C_8 = \frac{\color{blue}{9!}}{\color{red}{8!}\cdot\color{green}{ (9-8)!}}= \frac{\color{blue}{362880}}{\color{red}{40320}\cdot \color{green}{1}}= \color{brown}9\]
:P

ta-da XD

Answer 97

headache ;-; ... I think I got you though lol

Answer 98

My point.
If you want to use the shortcut I described, then you'd want to have as small an r as possible.

So... Say, when asked to solve 8C5

Just remember that this is the same as 8C3, and use this <---- instead, since 3 is smaller than 5 and less likely to cause you headaches.