uri
  • uri
How many different committies of 3 men and 4 girls can be formed from 8 men and 6 girls.
Mathematics
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SOLVED
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jamiebookeater
  • jamiebookeater
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shamim
  • shamim
\[C _{3}^{8} \times C _{4}^{6}\]
shamim
  • shamim
C= combination
uri
  • uri
Have you written the formula the way it should be? @shamim

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More answers

terenzreignz
  • terenzreignz
ohh combinations this time :D You have an idea what @shamim meant? :D This is \[\Large ^nC_r\] this time
shamim
  • shamim
ya
uri
  • uri
Yes! Got it.
terenzreignz
  • terenzreignz
Okay, admire this formula this time around :D \[\huge ^nC_r = \frac{n!}{(n-r)!\color{red}{r!}}\]
terenzreignz
  • terenzreignz
I put the r! at the bottom in red to emphasise its difference from nPr
shamim
  • shamim
ya i was meaning it
terenzreignz
  • terenzreignz
You DO remember what factorial means, right @uri ? :)
uri
  • uri
Yes i do remember :)
terenzreignz
  • terenzreignz
Good. That's all I want to know :P
uri
  • uri
But tell me how will they be written in the formula @terenzreignz
uri
  • uri
@msingh Can you please give me the link again? :3
terenzreignz
  • terenzreignz
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
uri
  • uri
Hey amistre explained me the difference yesterday :D @terenzreignz I know
terenzreignz
  • terenzreignz
all right, fine :/
terenzreignz
  • terenzreignz
then for the first one \[\Large ^8C_3 = \frac{8!}{(8-3)!3!}= \frac{8!}{5!\cdot 3!}\]
cwrw238
  • cwrw238
one way to work out combinations ( if your calculator doesn't help) is as follows 8C3 = 8 * 7 * 6 ------- = 56 3 * 2 * 1
cwrw238
  • cwrw238
8C2 = 8*7 --- = 28 2* 1
uri
  • uri
7C3 =7P3/3 =210/3! = :35 Just an example,We have to do it like that? @terenzreignz
terenzreignz
  • terenzreignz
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: \[\huge ^8C_3=\frac{8\cdot7\cdot6}{1\cdot2\cdot3}\]
cwrw238
  • cwrw238
right
cwrw238
  • cwrw238
also its worth knowing that 8C5 = 8C(8-5) = 8C3
cwrw238
  • cwrw238
7C6 = 7C1 = 7
uri
  • uri
8C3 =8P3/3! =336/3! =56 right? :D
cwrw238
  • cwrw238
yes
terenzreignz
  • terenzreignz
Yeah, on the whole, yes :D \[\huge ^nC_r = \frac{^nP_r}{r!}\]
uri
  • uri
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?
cwrw238
  • cwrw238
4! = 4*3*2*1
uri
  • uri
Got it :D
terenzreignz
  • terenzreignz
4! = 4 times 3! haha That applies to all positive integers by the way... \[\huge n! = n\times (n-1)!\]
uri
  • uri
6C4 =6P4/4! =360/4! =15..?
terenzreignz
  • terenzreignz
It's like slowly unraveling a tape or a roll of thread... \[\large 8!=8\cdot 7!\]\[\large 8!=8\cdot 7\cdot6!\]\[\large 8!=8\cdot 7\cdot6\cdot5!\]\[\large 8!=8\cdot 7\cdot6\cdot5\cdot4!\]
terenzreignz
  • terenzreignz
That is correct :)
uri
  • uri
OMG YAAAY @terenzreignz xD
terenzreignz
  • terenzreignz
:) Kudos @shamim @cwrw238 ^.^
uri
  • uri
Terence give @cwrw238 a medal foh meh :D
cwrw238
  • cwrw238
lol ty
uri
  • uri
:3

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