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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?

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 \(C_{10}^3 - C_6^3 = 120 - 20 = 100\) . The correct answer is E. -------------------------------------------------------------------------------- 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.

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.

-------------------------------------------------------------------------------- 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!! _________________

--------------------------------------------------------------------------------------- If you think you can,you can If you think you can't,you are right.

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

Answer is E. 100. Break this down to: 4C3 (All professors)+ 6C2 * 4C1 (2 students + 1 professor)+ 6C1*4C2 (2 professors + 1 student) =4+60+36 = 100 _________________

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?

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 _________________

Happy are those who dream dreams and are ready to pay the price to make them come true

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?

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 \(C_{10}^3 - C_6^3 = 120 - 20 = 100\) . The correct answer is E. -------------------------------------------------------------------------------- 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):

{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}:

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?

Thanks in advance

It is = (total possibilities) - (only students) - (2 students + 1 professor) = 10c3 - 6c3 - (4c1 * 6c2) _________________

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

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 ???