Question

How many different committies of 3 men and 4 girls can be formed from 8 men and 6 girls.

Answer 1

\[C _{3}^{8} \times C _{4}^{6}\]

Answer 2

C= combination

Answer 3

Have you written the formula the way it should be? @shamim

Answer 4

ohh combinations this time :D

You have an idea what @shamim meant? :D

This is
\[\Large ^nC_r\] this time

Answer 5

ya

Answer 6

Yes! Got it.

Answer 7

Okay, admire this formula this time around :D
\[\huge ^nC_r = \frac{n!}{(n-r)!\color{red}{r!}}\]

Answer 8

I put the r! at the bottom in red to emphasise its difference from nPr

Answer 9

ya i was meaning it

Answer 10

You DO remember what factorial means, right @uri ?
:)

Answer 11

Yes i do remember :)

Answer 12

Good. That's all I want to know :P

Answer 13

But tell me how will they be written in the formula @terenzreignz

Answer 14

@msingh Can you please give me the link again? :3

Answer 15

Let me explain the difference between nPr and nCr in a 'nice' manner.

Say you have 6 numbers
1 2 3 4 5 6

And you're asked how many THREE DIGIT NUMBERS you can form using these 6 numbers.
The answer would be nPr
examples would be
123
124
126
231

etc

Answer 16

Hey amistre explained me the difference yesterday :D @terenzreignz I know

Answer 17

all right, fine :/

Answer 18

then for the first one
\[\Large ^8C_3 = \frac{8!}{(8-3)!3!}= \frac{8!}{5!\cdot 3!}\]

Answer 19

one way to work out combinations ( if your calculator doesn't help) is as follows

8C3 = 8 * 7 * 6
------- = 56
3 * 2 * 1

Answer 20

8C2 = 8*7
--- = 28
2* 1

Answer 21

7C3 =7P3/3 =210/3! = :35 Just an example,We have to do it like that? @terenzreignz

Answer 22

Me and my formulas :D
Looks like @cwrw238 is showing us the simpler way to do it :)

if 8P3 has us picking three numbers, downwards (to 0) starting from 8...
8P3 = 8 x 7 x 6

8C3 would have a fraction, the numerator having us pick three numbers downwards starting from 8... while the denominator has us pick three numbers *upwards starting from 1*

Like so:
\[\huge ^8C_3=\frac{8\cdot7\cdot6}{1\cdot2\cdot3}\]

Answer 23

right

Answer 24

also its worth knowing that

8C5 = 8C(8-5) = 8C3

Answer 25

7C6 = 7C1 = 7

Answer 26

8C3 =8P3/3! =336/3! =56 right? :D

Answer 27

yes

Answer 28

Yeah, on the whole, yes :D
\[\huge ^nC_r = \frac{^nP_r}{r!}\]

Answer 29

And for the other one :
6C4 =6P4/4! =360/4!...right? But before that I have a Question....3! =6 right? so what does 4! mean?

Answer 30

4! = 4*3*2*1

Answer 31

Got it :D

Answer 32

4! = 4 times 3!
haha
That applies to all positive integers by the way...
\[\huge n! = n\times (n-1)!\]

Answer 33

6C4 =6P4/4! =360/4! =15..?

Answer 34

It's like slowly unraveling a tape or a roll of thread...
\[\large 8!=8\cdot 7!\]\[\large 8!=8\cdot 7\cdot6!\]\[\large 8!=8\cdot 7\cdot6\cdot5!\]\[\large 8!=8\cdot 7\cdot6\cdot5\cdot4!\]

Answer 35

That is correct :)

Answer 36

OMG YAAAY @terenzreignz xD

Answer 37

:)
Kudos @shamim @cwrw238 ^.^

Answer 38

Terence give @cwrw238 a medal foh meh :D

Answer 39

lol ty

Answer 40

:3