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# m25 q.3

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m25 q.3 [#permalink]  21 Feb 2009, 15:34
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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?

(A) 36
(B) 60
(C) 72
(D) 80
(E) 100

[Reveal] Spoiler: OA
E

Source: GMAT Club Tests - hardest GMAT questions

The best way to approach this problem is to consider an unconstrained version of the question first: how many committees of 3 are possible? The answer is $$C_{10}^3 = \frac{10!}{(7!3!)} = 120$$ . From this figure we have to subtract the number of committees that consist entirely of students i.e. $$C_{6}^3 = \frac{6!}{(3!3!)} = 20$$ . The final answer is $$C_{10}^3 - C_6^3 = 120 - 20 = 100$$ .
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I don't understand why we only subtract from the students only here? As first we take all the possible comibinations then minus just by if all the seats were to be filled by students only?? Why wont this only leave combinations of seat that will will be filled by just professors then? Dont understand the answer can someone please dumb it down to explain to me, many thanks JF.
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Re: m25 q.3 [#permalink]  22 Feb 2009, 12:43
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Here we subtract only those cases in which the committee has only student members because the question says "In how many ways can this committee be formed if it has to include at least 1 professor" so when we subtract the number of cases in which there are only students from total number of cases, we will get the number of cases in which the committee consists of atleast one professor. There is no upper limit to the number of professors that can be included in the committee so but there is a lower limit and hence we remove all the cases in which the condition for the lower limit is violated.

Hope I was able to solve your problem.
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Re: m25 q.3 [#permalink]  05 Jan 2010, 05:42
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I don't understand why we only subtract from the students only here? As first we take all the possible comibinations then minus just by if all the seats were to be filled by students only?? Why wont this only leave combinations of seat that will will be filled by just professors then? Dont understand the answer can someone please dumb it down to explain to me, many thanks JF.[/quote]

a simple approach is as follows:

There are
4P(professors) and 6S(Students)

A committee of 3 is to be chosen with at least 1 prof.
meaning:Committee can have (1P,2S) or (2P,1S) or (3P)---at least meaning 1 and more than 1...obv there cant more than 3 members in the commitee..
now calculating..
(1P,2S) - C(4,1)*C(6,2) =60
(2P,1S) - C(4,2)*C(6,1) =36
(3P) -C(4,3) =4

Add all the above(bcos all of the above are not possible simultaneously but only one of them at a time) and u get 100.

Happy to help!!
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Re: m25 q.3 [#permalink]  05 Jan 2010, 07:11
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there are 3 possibilities

i) all 3 professors
4c3 = 4
ii)2 professors & 1 student
4c2 * 6c1 = 6 *6 = 36
iii)1 professor & 2 strudent
4c1 * 6c2 = 4*15 =60

total possibilities = 4+36+60=100

so my choice is E
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Re: m25 q.3 [#permalink]  05 Jan 2010, 08:07
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There are 3 cases here (Because it says at least 1 professor):
1st Case: Choose only 1 professor and 2 student:
4c1 * 6c2= 60
2nd case: Choose 2 professors and 1 student:
4c2 * 6c1 = 36
3rd case: Choose 3 professors:
4c3 = 4

correct ans is E
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Re: m25 q.3 [#permalink]  05 Jan 2010, 15:37
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100.
Break this down to:
4C3 (All professors)+
6C2 * 4C1 (2 students + 1 professor)+
6C1*4C2 (2 professors + 1 student)
=4+60+36 = 100
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Re: m25 q.3 [#permalink]  05 Jan 2010, 19:39
Ans: E first find total no of possibale group 10C3 = 120

now find no of group thet can be form without single prof.(only student) 6C3=20

so the no of group in which atlast one prof. is 120-20=100
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Re: m25 q.3 [#permalink]  18 Apr 2010, 22:47
Thanks for all the explanations everyone, especially ddtiku.

I think this is a good example from which we can learn how to tackle many of the other possible twists they can spin on us in the gmat.

Here are a few examples that I would really appreciate some help on:

how would we calculate for at least 2 professors?

how would we calculate the probability the probablility that at least 2 professors would be selected as part of the committee if 1 definitely had to be selected?

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Re: m25 q.3 [#permalink]  04 Jan 2011, 23:52
Are the r and n in C (n,r) flip flopped in the explanation in the testing module?
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Re: m25 q.3 [#permalink]  11 Jun 2011, 08:18
arrangements with at least 1 professor in committee of 3 = total possible arrangements - no professors in committee
of 3
= 10C3 - 6c3 = 120 - 20 =100

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Re: m25 q.3 [#permalink]  13 Jan 2012, 19:08
Having problems with probability and combination.

Can somebody tell me what's wrong with this methodology?
at least 1 prof - 4x6x5 = 120
at least 2 profs - 4x3x6 = 72
at least 3 profs - 4x3x2 = 24

Sum = 120+72+24= 216
divide by 3 to remove the repeats = 72.

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Re: m25 q.3 [#permalink]  15 Jan 2012, 04:53
mourinhogmat1 wrote:
Having problems with probability and combination.

Can somebody tell me what's wrong with this methodology?

ur problem is that u have counted the same ppl several times. u should divide them by 2! or 3!
at least 1 prof - 4x6x5/2! = 120/2=60
at least 2 profs - 4x3x6 = 72/2!=36
at least 3 profs - 4x3x2 = 24/3!=4

60+36+4=100

the 2nd approach -

at least 1 prof - 4C1*6C2=60
at least 2 profs 4C2*6C1=36
at least 3 profs 4C3=4

hope it helps
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Re: m25 q.3 [#permalink]  15 Jan 2012, 22:07
Ans is E.
Since we need at least one professor so the combinations can be PSS+PPS+PPP = 4c1*6c2+4c2*6c1+4c3 = 100
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Re: m25 q.3 [#permalink]  16 Jan 2012, 03:19
Expert's post
1
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Forrester300 wrote:
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?

(A) 36
(B) 60
(C) 72
(D) 80
(E) 100

[Reveal] Spoiler: OA
E

Source: GMAT Club Tests - hardest GMAT questions

The best way to approach this problem is to consider an unconstrained version of the question first: how many committees of 3 are possible? The answer is $$C_{10}^3 = \frac{10!}{(7!3!)} = 120$$ . From this figure we have to subtract the number of committees that consist entirely of students i.e. $$C_{6}^3 = \frac{6!}{(3!3!)} = 20$$ . The final answer is $$C_{10}^3 - C_6^3 = 120 - 20 = 100$$ .
--------------------------------------------------------------------------------
I don't understand why we only subtract from the students only here? As first we take all the possible comibinations then minus just by if all the seats were to be filled by students only?? Why wont this only leave combinations of seat that will will be filled by just professors then? Dont understand the answer can someone please dumb it down to explain to me, many thanks JF.

4 professors and 6 students are being considered for membership on a supervisory committee which must consist of 3 people. If the committee has to include at least 1 professor, how many ways can this committee be formed?
A. 36
B. 60
C. 72
D. 80
E. 100

{The committees with at least one professor} = {Total committees possible} - {The committees with zero professors} (so minus the committees with only students in them).

So, $$C^3_{10}$$ (total # of selection of 3 out of 10) minus $$C^3_6$$ (# of selection of 3 person from 6 students, which means zero professor):

$$C^3_{10}-C^3_6=\frac{10}{7!*3!}-\frac{6!}{3!*3!}=120-20=100$$.

Or direct approach:

{The committees with at least one professor} = {The committees with 1 professor / 2 students} + {The committees with 2 professors / 1 student} + {The committees with 3 professors / 0 students}:

OR: $$C^1_4*C^2_6+C^2_4*C^1_6+C^3_4=100$$;

Also discussed here: why-am-i-wrong-85865.html
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Re: m25 q.3 [#permalink]  21 Feb 2012, 06:01
cyrusthegreat wrote:
Thanks for all the explanations everyone, especially ddtiku.

I think this is a good example from which we can learn how to tackle many of the other possible twists they can spin on us in the gmat.

Here are a few examples that I would really appreciate some help on:

how would we calculate for at least 2 professors?

how would we calculate the probability the probablility that at least 2 professors would be selected as part of the committee if 1 definitely had to be selected?

It is = (total possibilities) - (only students) - (2 students + 1 professor)
= 10c3 - 6c3 - (4c1 * 6c2)
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Re: m25 q.3 [#permalink]  22 Feb 2012, 10:46
I was running into similar issue. Thanks for all the explanations.
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Re: m25 q.3 [#permalink]  14 Mar 2012, 01:26
these are the type of questions that trip me up... i thought that the committee is 1 professor and 2 students
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Re: m25 q.3 [#permalink]  15 Jan 2013, 05:24
Total ways = 120 ways (as explained above).
Probability that all seats are for students = 6/10X5/9X4/8 = 1/6
Nos of ways that there are 3 students in all three seats = 1/6X120 = 20 ways
Nos of ways that there is atleast 1 proffesor on any one of three seats = 120-20 = 100 ways
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Re: m25 q.3 [#permalink]  15 Jan 2013, 08:13
This was among hardest questions of GMAT or gmatclub?
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Re: m25 q.3 [#permalink]  03 Jan 2014, 10:03
Y cant we calculate as follows
No of ways of selecting a professor is 4C1
No of ways of selecting other 2 members 9C2
Total number of ways = 4C1×9C2 ???
Re: m25 q.3   [#permalink] 03 Jan 2014, 10:03

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