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An engagement team consists of a project manager, team leade [#permalink]
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31 Jan 2008, 22:09
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An engagement team consists of a project manager, team leader, and four consultants. There are 2 candidates for the position of project manager, 3 candidates for the position of team leader, and 7 candidates for the 4 consultant slots. If 2 out of 7 consultants refuse to be on the same team, how many different teams are possible? A. 25 B. 35 C. 150 D. 210 E. 300
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Last edited by Bunuel on 19 Feb 2014, 00:07, edited 1 time in total.
Renamed the topic, edited the question, added the OA and moved to PS forum.



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Re: combo with constraints [#permalink]
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31 Jan 2008, 22:47
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marcodonzelli wrote: An engagement team consists of a project manager, team leader and four consultants. There are 2 candidates for the position of proj.manager, 3 for the position of team leader and 7 for 4 consultants slots. If 2 of the 7 consultants refuse to be on the same team, how many different team are possible?
25 35 150 210 300 I get C. If I see this on the test, id prolly get to 210 and then guess A B or C. OK so for the first position we have only 2 possiblities ( proj manager) 3 for the team leader so and 7!/3!4! for the consultants 2*3*35 > 210 Now I dunno how to figure out the constraints quickly but I eventually figured it out. Obvs. we want to figure out the total ways in which the two ARE on the same team. I did it by AB are the two and XYZFN are the rest ABXO (O stands for the other 4) So the first is ABXY, Z,F,N 4 possible choices The next is ABYO (but notice we already had YX so there are only 3 possible choices) ABZO ABFO ABNO No possible here we used them all up So 4+3+2+1 = 10 SO now its 2*3*10 =60 So 21060 = 150



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Re: combo with constraints [#permalink]
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01 Feb 2008, 10:15
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Following you up till the constraint part...Looking for a combo master to teach us how to do it by calculation and not by hand



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Re: combo with constraints [#permalink]
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01 Feb 2008, 10:25
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marcodonzelli wrote: An engagement team consists of a project manager, team leader and four consultants. There are 2 candidates for the position of proj.manager, 3 for the position of team leader and 7 for 4 consultants slots. If 2 of the 7 consultants refuse to be on the same team, how many different team are possible?
25 35 150 210 300 2C1*3C1*(5C4+5C3*2C1)=2*3*(5+10*2)=6*25=150 > C



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Re: combo with constraints [#permalink]
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01 Feb 2008, 10:27
Can you explain this part is words:
(5C4+5C3*2C1)



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Re: combo with constraints [#permalink]
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01 Feb 2008, 10:35
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jimmyjamesdonkey wrote: Can you explain this part is words:
(5C4+5C3*2C1) yes, can you explain?



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Re: combo with constraints [#permalink]
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01 Feb 2008, 10:42
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jimmyjamesdonkey wrote: Can you explain this part is words:
(5C4+5C3*2C1) We can either choose 4 consultants from a group of 5 (5C4) who are willing to work with everyone or choose 3 from that group (5C3) and one person from the other group of 2 consultants (2C1) who don't want to work with each other.



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Re: combo with constraints [#permalink]
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01 Feb 2008, 10:55
Very cool. Thanks. +1



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Re: combo with constraints [#permalink]
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02 Feb 2008, 06:20
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To get the number of ways you can select 4 consultants, you can do this:
A. Total number of ways 4 can be selected out of 7 = 7C4 = 35 B. Total number of ways in which these two snobbish consultants are together (two are already there, so you just need to select rest two out of 5) = 5C2 = 10
A (minus) B results in 25 which can then be multiplied with (2C1*3C1) to result in 150



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Re: combo with constraints [#permalink]
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02 Feb 2008, 12:20
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2C1*3C1*7C4  2C1*3C1*5C2 = 6[3510] = 150.
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Re: An engagement team consists of a project manager, team leade [#permalink]
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10 Mar 2014, 23:34
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Answer= Total number combinations  Total number of combinations with constraints Total number of combinations = 2C1*3C1*7C4= 210 Total number of combinations with constraints = 2C1*3C1*5C2=60 Answer=21060=150
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Re: An engagement team consists of a project manager, team leade [#permalink]
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14 May 2014, 03:46
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Answer: C a) No of ways to select 1 Manager = 2c1 = 2 b) No of ways to select 1 Team leader = 3c1 = 3 c) No of ways to select 4 Consultants = 7c4 = 35 Therefore, possible teams without any constraint = 2x3x35 = 210
No of ways to select 4 Consultants out of 7 when 2 of them are always together = 6c4 x2! = 60
Therefore, possible teams with given constraint = 210  60 = 150
Hope its clear.



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Re: An engagement team consists of a project manager, team leade [#permalink]
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17 May 2014, 02:25
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Well, probably the quickest way to do this problem is to
1) eliminate D & E, because we are pretty sure it is less than 210 (we know max is 2*3*35). 2) eliminate A & B, because we know 25 and 35 are not multiple of 6 (we know the combination will be an integer).



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An engagement team consists of a project manager, team leade [#permalink]
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21 Nov 2015, 12:21
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srp wrote: To get the number of ways you can select 4 consultants, you can do this:
A. Total number of ways 4 can be selected out of 7 = 7C4 = 35 B. Total number of ways in which these two snobbish consultants are together (two are already there, so you just need to select rest two out of 5) = 5C2 = 10
A (minus) B results in 25 which can then be multiplied with (2C1*3C1) to result in 150 Simple and straightforward! (Also, I used this method as well! ) I'd like to add that your B is actually 2C2 * 5C2 = 1 * 10 = 10 We have 2C2, because we have already chosen the consultants in the group.



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Re: An engagement team consists of a project manager, team leade [#permalink]
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13 Jul 2016, 05:05
marcodonzelli wrote: An engagement team consists of a project manager, team leader, and four consultants. There are 2 candidates for the position of project manager, 3 candidates for the position of team leader, and 7 candidates for the 4 consultant slots. If 2 out of 7 consultants refuse to be on the same team, how many different teams are possible?
A. 25 B. 35 C. 150 D. 210 E. 300 This is how I solved this question and got the answer incorrect. Total Combinations  (the number of combinations in which A and B are always together) = Number of ways in which A and B won't work together Assuming that A and B are the two consultants who don't want to work together. Number of ways to choose 1 PM out of 2 candidates x Number of ways to choose 1 TL out of 3 candidates x Number of ways to choose 4 Consultants out of 7 candidates = Total 2 ways x 3 ways x 35 ways(7!/4!.3!) = 210 ways. I got this right. However, somewhere in the next steps is where I made a mistake. Number of ways in which A and B will work together and the rest two positions can be filled with the remaining 5 candidates. 2 x 3 x 1 x 1 x 5 x 4 So I subtracted 120 from 210. However after following the answers here, I understood that for the remaining two positions, instead of doing a 5C2, I chose to fill them in 5 ways and 4 ways. I would like to understand: 1) what sort of an error am I committing. 2When should I follow the approach I used above and when should I do a 5C2 logic. Suggestions welcome. Thanks! Regards, Arvind.



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Re: An engagement team consists of a project manager, team leade [#permalink]
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13 Jul 2016, 23:42
total teams 2c1 * 3c1* 7c4 = 210 let both be on same team and always selected so now we have to choose 2 out of remaining 5 total team when both candidates on same team will be 2c1 * 3c1* 5c2 = 60 21060 = 150



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An engagement team consists of a project manager, team leade [#permalink]
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14 Jul 2016, 08:02
Hi All, The answer choices to this question provide us with an interesting 'shortcut' that we can use to avoid some of the math involved. Since there are 2 possible project managers and 3 potential team leaders, then the final answer MUST be a multiple of (2)(3) = 6. Since we're choosing 4 of 7 possible consultants, we can use the Combination Formula: 7!/(4!3!) = 35 possible groups of 4 consultants. IF there were no additional restrictions, then there would be (6)(35) = 210 possible groups. HOWEVER, we know that certain consultants won't work with other consultants, so the number of possible groups must be LESS than 210. Based on the answer choices, there's only one option that is a multiple of 6 and is less than 210... Final Answer: GMAT assassins aren't born, they're made, Rich
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Re: An engagement team consists of a project manager, team leade [#permalink]
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22 Dec 2016, 01:34
marcodonzelli wrote: An engagement team consists of a project manager, team leader, and four consultants. There are 2 candidates for the position of project manager, 3 candidates for the position of team leader, and 7 candidates for the 4 consultant slots. If 2 out of 7 consultants refuse to be on the same team, how many different teams are possible?
A. 25 B. 35 C. 150 D. 210 E. 300 i did this question in this way 2c1 * 3c1* 5c4+5c3*2c1= 2*3*(5+5*4*2/2)=6*25=150



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Re: An engagement team consists of a project manager, team leade [#permalink]
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30 Aug 2017, 03:35
A very easy approach : Total ways without the constraints = 2C1 x 3C1 x 7C4 = 2 * 3 * 7C4 = 6 * 7C4 = 210. Now we know that the answer is less than 210 and a multiple of 6. So answer = C. Cheers



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Re: An engagement team consists of a project manager, team leade [#permalink]
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02 Sep 2017, 07:10
marcodonzelli wrote: An engagement team consists of a project manager, team leader, and four consultants. There are 2 candidates for the position of project manager, 3 candidates for the position of team leader, and 7 candidates for the 4 consultant slots. If 2 out of 7 consultants refuse to be on the same team, how many different teams are possible?
A. 25 B. 35 C. 150 D. 210 E. 300 We can select the project manager in 2 ways and the team leader in 3 ways. For the consultants, there are 7 candidates for 4 positions, but 2 of the 7 cannot be together. We can use the following equation: # of ways to select the consultants = # of ways with the 2 together + # of ways with the 2 not together. The total number of ways to select the consultants is 7C4 = 7!/[4!(74)!] = 7!/(4!3!) = (7 x 6 x 5 x 4)/4! = 7 x 6 x 5 x 4)/(4 x 3 x 2) = 35. The number of ways to select the consultants when the 2 are together can be found by choosing 2 people for the remaining slots from the 5 remaining candidates, which is given by 5C2 = (5 x 4)/2 = 10. Thus, the number of ways to select the consultants when the 2 are not together is 35  10 = 25. So, the total number of ways to select the teams is 2 x 3 x 25 = 150. Answer: C
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