4 professors and 6 students

[DavoodBeater]

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
[/spoiler]

[logitech]

AT LEAST 1 PROF = ALL POSSIBILITIES - ALL STUDENTS

= 10C3 - 6C3
= 100

[DavoodBeater]

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.

[logitech]

How about this:

1 Pro + 2 Pro + 3 Pro = 4c1x6c2 + 4c2x6c1 + 4c3 = 60 + 36 + 4 = 100

[dubeystuts]

I am sorry but I do understand how u get 100. Can someone please do ahuge favor by explaining this again? Many Thanks.

[dubeystuts]

I am sorry but I do understand how u get 100. Can someone please do ahuge favor by explaining this again? Many Thanks.

[DavoodBeater]

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.

[logitech]

Time for fundementals!. Dont try to run before you learn how to walk

[tongtoey]

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)!

[welcome]

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.

[DavoodBeater]

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.

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]

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 !!!

so you are saying

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

[DavoodBeater]

no my friend,
by (4 1) I mean 4C1, for instance.
It is clear now.
Thanks everybody.

[logitech]

no my friend,
by (4 1) I mean 4C1, for instance.
It is clear now.
Thanks everybody.

Believe me I know what you mean by (4 1)