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### combo

by gmatusa2010 » Thu Dec 23, 2010 11:29 pm
A certain academic department consists of 3 senior professors
and 6 junior professors. How many diÂ¤erent committees of 3
professors can be formed in which at least one member of the
committee is a senior professor? (Two groups are considered
diÂ¤erent if at least one group member is diÂ¤erent.)
(A) 168
(B) 127
(C) 66
(D) 64
(E) 36

What is wrong with my method:

1) At least one of the 3 has to have a SP. So Your 1st seat you have only 3 choices

2) Now you have 2 SP and 6 JP left. to arrange them in two seats 8!/2!6!=28

3) 3*28= 84.

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by Rezinka » Fri Dec 24, 2010 12:30 am
IMO : D
3 combinations are possible : 1S+2J, 2S+1J, 3S+0J
Taking the combinations for the above :
(3C1*6C2) + (3C2*6C1) + (3C3*6C0)
=45 + 18 + 1
=64

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by captcha » Fri Dec 24, 2010 12:57 am
A certain academic department consists of 3 senior professors
and 6 junior professors. How many diÃ‚Â¤erent committees of 3
professors can be formed in which at least one member of the
committee is a senior professor? (Two groups are considered
diÃ‚Â¤erent if at least one group member is diÃ‚Â¤erent.)

3 SP & 6 JP ---------> committee of 3 prof with atleast 1 SP

1 SP & 2JP + 2 SP & 1 JP + 3 SP & 0JP

=3C1*6C2 + 3C2*6C1 + 3C3*6C0
=3*15+3*6+1
=64

Other Way:

Total no. of groups possible - Groups with all 3 JPs

9C3 - 6C3

=84 - 20
=64

Attn: gmatusa2010

When doing 3C1 * 8C2

The groups formed are repeating eg.

SP={S1,S2,S3} & JP = {J1,J2,J3,J4,J5,J6}

3C1 * 8C2

=[S1;S2,J1] , [S2,S1,J1], [S1,S3,J1] .... and so on...

you will see that S1,S2,S3 will repeat in some combinations (20 to be precise )

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by anshumishra » Fri Dec 24, 2010 6:57 am
gmatusa2010 wrote:A certain academic department consists of 3 senior professors
and 6 junior professors. How many diÂ¤erent committees of 3
professors can be formed in which at least one member of the
committee is a senior professor? (Two groups are considered
diÂ¤erent if at least one group member is diÂ¤erent.)
(A) 168
(B) 127
(C) 66
(D) 64
(E) 36

What is wrong with my method:

1) At least one of the 3 has to have a SP. So Your 1st seat you have only 3 choices

2) Now you have 2 SP and 6 JP left. to arrange them in two seats 8!/2!6!=28

3) 3*28= 84.
captcha has provided two good solutions and also pointed out that your answer has some repeated combinations.

Whenever you get something to select don't think about arrangements (which I have made bold in your approach). The arrangement will make some repetitions here.

So, The group of people selected (S1, S2, J1) is same as (S2, S1, J1 ) or (J1, S1, S2)...
The arrangements make sense in the case where the position of the object selected makes difference (e.g. If you select 3 different keys of a number lock... the position of different digits makes the final key different).

Hopefully, that helps a bit.
Thanks
Anshu

(Every mistake is a lesson learned )

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