can some one explain why i am wrong?!
4 professors and 6 students are being considered for membership on a supervisory committee which must consist of 3 people. In how many ways can this committee be formed if it has to include at least 1 professor?
36
60
72
80
100
my answer:
chose one from 4 professors first, and then the rest of combination:
(4 1) * [ (3 1)*(6 1) + (3 2) + (6 2)] = 144 !!!
OA is 100
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4 professors and 6 students
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thanks Logitech, indeed i know that way.
but i wanted to know my mistake.
and i think it was that i multiplied the first (4 1) to everyone which is wrong and the approach was totally wrong in this case. It is useful for probability, not counting.
but i wanted to know my mistake.
and i think it was that i multiplied the first (4 1) to everyone which is wrong and the approach was totally wrong in this case. It is useful for probability, not counting.
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I am sorry but I do understand how u get 100. Can someone please do ahuge favor by explaining this again? Many Thanks.
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I am sorry but I do understand how u get 100. Can someone please do ahuge favor by explaining this again? Many Thanks.
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the easiest way is in post 2:
all - no professor at all (or you can say all student)
=> 10C3 - 6C3 = 100
the second way is in post 4:
it means count number of way that there is only one prof, count when there are 2 and count when there are 3 prof.
=> 4c1x6c2 + 4c2x6c1 + 4c3 = 100
first one mean 1 prof and 2 students, second means 2 prof. and 1 student, and third means 3 prof.
but the toughest way is my way! i done even understand that! it is not correct, i mean.
all - no professor at all (or you can say all student)
=> 10C3 - 6C3 = 100
the second way is in post 4:
it means count number of way that there is only one prof, count when there are 2 and count when there are 3 prof.
=> 4c1x6c2 + 4c2x6c1 + 4c3 = 100
first one mean 1 prof and 2 students, second means 2 prof. and 1 student, and third means 3 prof.
but the toughest way is my way! i done even understand that! it is not correct, i mean.
DavoodBeater,
The reason that your method
(4 1) * [ (3 1)*(6 1) + (3 2) + (6 2)] = 144 !!!
got more than the actual answer is because your method cosiders the order by reserving the first place for the first professor.(4 1)
So you might end up with choosing the same group of people more than once
Ex:
Professor: A, B, C, D
Student : Sn = S1, S2, S3, S4, S5, S6
For a group of 2 professor + 1 student
** underlined letter represents the first professor **
A + [ B + Sn ] = ABSn
A + [ C + Sn ] = ACSn
A + [ D + Sn ] = ADSn
B + [ A + Sn ] = BASn <--- or ABSn same as above
B + [ C + Sn ] = BCSn
B + [ D + Sn ] = BDSn
C + .... and so on
------------------------------------------------------------------
To pick X out of Y, order matters
Y! / (Y-X)!
which always more than
To pick X out of Y, order doesn't matter
Y! / X!(Y-X)!
The reason that your method
(4 1) * [ (3 1)*(6 1) + (3 2) + (6 2)] = 144 !!!
got more than the actual answer is because your method cosiders the order by reserving the first place for the first professor.(4 1)
So you might end up with choosing the same group of people more than once
Ex:
Professor: A, B, C, D
Student : Sn = S1, S2, S3, S4, S5, S6
For a group of 2 professor + 1 student
** underlined letter represents the first professor **
A + [ B + Sn ] = ABSn
A + [ C + Sn ] = ACSn
A + [ D + Sn ] = ADSn
B + [ A + Sn ] = BASn <--- or ABSn same as above
B + [ C + Sn ] = BCSn
B + [ D + Sn ] = BDSn
C + .... and so on
------------------------------------------------------------------
To pick X out of Y, order matters
Y! / (Y-X)!
which always more than
To pick X out of Y, order doesn't matter
Y! / X!(Y-X)!
Last edited by tongtoey on Thu Jan 08, 2009 10:11 am, edited 1 time in total.
Total 4 P + 6 P = 10
members in committe = 3
Total ways 10C3 = 10!/(7!*3!) = 120
These 120 ways includes, those with no prof.
So if we reduse those ways, we will get the ways by which we can make committe with at lease 1 prof.
Ways with no profs = 6C3 = 20.
Answer = 120-20 = 100.
members in committe = 3
Total ways 10C3 = 10!/(7!*3!) = 120
These 120 ways includes, those with no prof.
So if we reduse those ways, we will get the ways by which we can make committe with at lease 1 prof.
Ways with no profs = 6C3 = 20.
Answer = 120-20 = 100.
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thanks man, I am now more clear.
however i feel that it is possible to a solve the problem like this with a method like this,
but i dont exactly know how
thanks.
however i feel that it is possible to a solve the problem like this with a method like this,
but i dont exactly know how
thanks.
tongtoey wrote:DavoodBeater,
The reason that your method
(4 1) * [ (3 1)*(6 1) + (3 2) + (6 2)] = 144 !!!
got more than the actual answer is because your method cosiders the order by reserving the first place for the first professor.(4 1)
So you might end up with choosing the same group of people more than once
Ex:
Professor: A, B, C, D
Student : Sn = S1, S2, S3, S4, S5, S6
For a group of 2 professor + 1 student
** underlined letter represents the first professor **
A + [ B + Sn ] = ABSn
A + [ C + Sn ] = ACSn
A + [ D + Sn ] = ADSn
B + [ A + Sn ] = BASn <--- or ABSn same as above
B + [ C + Sn ] = BCSn
B + [ D + Sn ] = BDSn
C + .... and so on
------------------------------------------------------------------
To pick X out of Y, order matters
Y! / (Y-X)!
which always more than
To pick X out of Y, order doesn't matter
Y! / X!(Y-X)!
- logitech
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so you are sayingmy answer:
chose one from 4 professors first, and then the rest of combination:
(4 1) * [ (3 1)*(6 1) + (3 2) + (6 2)] = 144 !!!
41x31x61 + 41x32 + 41x62
Here is where your approach fails:
for example
lets take the middle part
(4 1) x ( 3 2)
This is not equal to the (4 3)
(4 1) x ( 3 2) = 4 x 3 = 12
but ( 4 3 ) = 4
WHY ?
the first one is listing the same GROUPS for 3 times
in other words
AB and BA are the same group
and you count them 1 not 2
Hope it gets more clear for you
LGTCH
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"DON'T LET ANYONE STEAL YOUR DREAM!"
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"DON'T LET ANYONE STEAL YOUR DREAM!"
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- logitech
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Believe me I know what you mean by (4 1)DavoodBeater wrote:no my friend,
by (4 1) I mean 4C1, for instance.
It is clear now.
Thanks everybody.
LGTCH
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"DON'T LET ANYONE STEAL YOUR DREAM!"
---------------------
"DON'T LET ANYONE STEAL YOUR DREAM!"