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# Permutations & Combinations : Help is on the way!

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Re: Permutations & Combinations : Help is on the way! [#permalink]

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04 Jun 2009, 07:56
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Re: Combination question baffling my mind [#permalink]

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14 Nov 2009, 05:48
bondguy wrote:
Came across this question on this site itself and I am struggling to get the answer the way they describe it and why my approach is at fault

here is the question:

EXAMPLE 10 A three-person committee must be chosen from a group of 7 professors and 10 graduate students. If at least one of the people on the committee must be a professor, how many different groups of people could be chosen for the committee?

A. 70
B. 560
C. 630
D. 1,260
E. 1,980

My approach :

Since we need to have a professor for the 3-member committee , we can choose the first person as professor in 7 ways .
Rest two positions can be filled with remaining professors or students in (16 C 2 )

total number of combinations for filling 3 member committe -
first position with professor * rest with 2 from 16 mombers ( 6 professors + 10 students) is
7 * (16 C 2) = 7 * (16 * 15 )/2
= 840

but what is baffling is total number of combinations for choosing 3 members from 17 ( 7 professors + 10 students) without restriction of atleast one professor in the committie = 17 C 2 = 680

how can this combination be less than restricted combination

can someone explain me where my thought process took a wrong turn ..

Here is how I solved this problem...

There are three cases in which you have a professor that is part of the comitte:
1 prof, 2 students
2 prof, 1 students
3 prof, 0 students

Here are the total number of combinations I found for each case:
1 prof, 2 students: 7*10*9 = 630
2 prof, 1 students: 7*6*10 = 420
3 prof, 0 students: 7*6*5 = 210

So then I got 1260 total (630+420+210) which is D.
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Re: Permutations & Combinations : Help is on the way! [#permalink]

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14 Nov 2009, 05:58
Woops, just realized what I did wrong: (I did a permutation instead of a combination)

The total number of combinations for each case should be:
1 prof, 2 students: 7C1 + 10C2
2 prof, 1 students: 7C2 + 10C1
3 prof, 0 students: 7C3

Total: 560

The easier way to do it would have been to find:

Total number of combinations - Number of combinations WITHOUT a professor:
17C3 - 10C3
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Re: Permutations & Combinations : Help is on the way! [#permalink]

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16 Nov 2009, 01:55
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Re: Permutations & Combinations : Help is on the way! [#permalink]

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31 Dec 2013, 13:00
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: Permutations & Combinations : Help is on the way! [#permalink]

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30 May 2015, 03:45
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: Permutations & Combinations : Help is on the way! [#permalink]

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02 Jun 2016, 01:06
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: Permutations & Combinations : Help is on the way! [#permalink]

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02 Jun 2016, 01:56
walletless wrote:

The advanced links explain the concept of P & C very well.
Re: Permutations & Combinations : Help is on the way!   [#permalink] 02 Jun 2016, 01:56

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# Permutations & Combinations : Help is on the way!

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