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A team of 6 cooks is chosen from 8 men and 5 women. The team [#permalink]
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16 Nov 2012, 07:38
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A team of 6 cooks is chosen from 8 men and 5 women. The team must have at least 2 men and at least 3 women. How many ways can this team be created? A) 140 B) 320 C) 560 D) 700 E) 840 For those of you studying Combinations and Probability, we've posted a free lesson on combinations here: http://www.gmatpill.com/gmatpracticet ... onslessonThe video explanation for the above question is at 22:13 Part 1: Combination Framework  Dice, Marbles, Pocket Pair" Part 2: Apply to GMAT  Word Problems Part 3: Poker Probability  Pocket Pair vs Pocket Aces, Flush/Full House 00:00  Intro 01:00  Warmup 02:12  Part 1 Overview: Combinations & Permutations vs Variations 08:35  Part 1 Details: Picking Teams, Dice, Pocket Pair, Dating 17:21  Part 2: Apply to GMAT  Word Problem #1 22:13  Part 2: Apply to GMAT  Word Problem #2 31:34  Part 2: Apply to GMAT  Word Problem #3 38:04  Part 3: Poker Probability  Pocket Pair vs Pocket Aces 43:40  Part 3: Poker Probability  Given Pocket Aces, 3ofaKind? 46:13  Part 3: Poker Probability  1Pair by 5th Flop 49:59  Part 3: Poker Probability  Full House/Flush 1:00:00  Conclusion
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Re: Cooking Team Combinations [#permalink]
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16 Nov 2012, 08:14
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gmatpill wrote: A team of 6 cooks is chosen from 8 men and 5 women. The team must have at least 2 men and at least 3 women. How many ways can this team be created?
A) 140 B) 320 C) 560 D) 700 E) 840
Only possible combinations are a team of 2M, 4 W or 3M,3W. Possible ways to make a team of 2M,4W = 8C2 * 5C4 =28*5 =140 Possible ways to make a team of 3M,3W = 8C3* 5C3 = 56*10 = 560 Total possible ways = 140+560 = 700 Ans D it is.
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Re: Cooking Team Combinations [#permalink]
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24 Dec 2012, 02:32
Could someone point out where my thought process is faltering. I did \(8C2 * 5C3 * 8C1\) 8C1  after choosing 2 men and and 3 women it doesn't matter who you pick from the remaining 8, so \(8C1\) Await valued response.
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Re: Cooking Team Combinations [#permalink]
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24 Dec 2012, 02:49
eaakbari wrote: Could someone point out where my thought process is faltering.
I did \(8C2 * 5C3 * 8C1\)
8C1  after choosing 2 men and and 3 women it doesn't matter who you pick from the remaining 8, so \(8C1\)
Await valued response. This way you are counting some teams more than once. Consider that you get A and B (men) from 8C2, X, Y, Z (women) from 5C3 and C (man) from 8C1, so your group is {A, B, C, X, Y, Z}. But if you get A and C (men) from 8C2, X, Y, Z (women) from 5C3 and B (man) from 8C1, your group will still be {A, B, C, X, Y, Z}. Hope it's clear.
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Re: Cooking Team Combinations [#permalink]
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24 Dec 2012, 03:02
Bunuel wrote: eaakbari wrote: Could someone point out where my thought process is faltering.
I did \(8C2 * 5C3 * 8C1\)
8C1  after choosing 2 men and and 3 women it doesn't matter who you pick from the remaining 8, so \(8C1\)
Await valued response. This way you are counting some teams more than once. Consider that you get A and B (men) from 8C2, X, Y, Z (women) from 5C3 and C (man) from 8C1, so your group is {A, B, C, X, Y, Z}. But if you get A and C (men) from 8C2, X, Y, Z (women) from 5C3 and B (man) from 8C1, your group will still be {A, B, C, X, Y, Z}. Hope it's clear. Ahhh, yes. I do understand now. Thanks, Bunuel, for the wonderful explanation.
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Re: A team of 6 cooks is chosen from 8 men and 5 women. The team [#permalink]
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25 Dec 2012, 21:08



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Re: A team of 6 cooks is chosen from 8 men and 5 women. The team [#permalink]
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26 Dec 2012, 03:26



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Re: A team of 6 cooks is chosen from 8 men and 5 women. The team [#permalink]
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26 Dec 2012, 05:22
Bunuel wrote: debayan222 wrote: OA: D700. GMATPill/Bunuel, Is it really a 700+ level Q ? It's pretty easy guys...! In that case combinations going to be an easy topic in GMAT.. GMAT combination/probability questions are fairly straightforward, so you won't see much harder questions on these topics on the exam. Great to know that Bunuel...! Well, in that case Qs. in your signature and those in the GMAT Club tests (obviously apart from OG Qs.) would be suffice for 700+ in GMAT I think as far as combination/probability questions are concerned...! Appreciate your thoughts on this...
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Re: A team of 6 cooks is chosen from 8 men and 5 women. The team [#permalink]
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27 Dec 2012, 22:56
gmatpill wrote: A team of 6 cooks is chosen from 8 men and 5 women. The team must have at least 2 men and at least 3 women. How many ways can this team be created?
A) 140 B) 320 C) 560 D) 700 E) 840
M M W W W M/W This means we could have either combinations: (1) 2MEN 4WOMEN (2) 3MEN 3WOMEN We will add the total number of both possibilities: 8!/6!2! * 5!/4!1! + 8!/3!5! * 5!/3!2! = 140 + 560 = 700 Answer: D
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A team of 6 cooks is chosen from 8 Men and 5 Women [#permalink]
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01 Sep 2013, 19:06
A team of 6 cooks is chosen from 8 Men and 5 Women. The team must have at least 2 men and at least 3 woen. How many ways can we form this team?
A) 140 B) 320 C) 560 D) 700 E) 840
Solution: We have two possibilities 2M 4F or 3M 3F (8C2)(5C3)+(8C3)(5C3)=700
I am trying another method but I am not getting it correctly. So for The first scenario 1) 3M 4F MMFFFF We can use a permutation approach. For the first man we have 8 choices. Second man 7. Likewise for the females. 5 for the first....2 for the last. We get (8*7)*(5*4*3*2) Now here is what is confusing me. Since I have MMFFFF and order doesn't matter the number of ways to arrange 3M's & 4F's is 6!/(2!*4!). I know I have to divide by 2! and 4! because we have duplicates and order doesn't matter but what about the 6!??? Should we include or exclude it? The answer has it excluded, but why?? I feel without the 6! we don't take into account MFFFMF or FFMFMF.... etc. What am I doing wrong?



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Re: A team of 6 cooks is chosen from 8 Men and 5 Women [#permalink]
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28 Sep 2013, 02:11
(8C2)(5C3)+(8C3)(5C3)=28*10 + 56*10 = 280 + 560 = 840 > E
are you sure its D? Please explain



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Re: A team of 6 cooks is chosen from 8 Men and 5 Women [#permalink]
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28 Sep 2013, 02:54
alphabeta1234 wrote: A team of 6 cooks is chosen from 8 Men and 5 Women. The team must have at least 2 men and at least 3 woen. How many ways can we form this team?
A) 140 B) 320 C) 560 D) 700 E) 840
Solution: We have two possibilities 2M 4F or 3M 3F (8C2)(5C3)+(8C3)(5C3)=700
I am trying another method but I am not getting it correctly. So for The first scenario 1) 3M 4F MMFFFF We can use a permutation approach. For the first man we have 8 choices. Second man 7. Likewise for the females. 5 for the first....2 for the last. We get (8*7)*(5*4*3*2) Now here is what is confusing me. Since I have MMFFFF and order doesn't matter the number of ways to arrange 3M's & 4F's is 6!/(2!*4!). I know I have to divide by 2! and 4! because we have duplicates and order doesn't matter but what about the 6!??? Should we include or exclude it? The answer has it excluded, but why?? I feel without the 6! we don't take into account MFFFMF or FFMFMF.... etc. What am I doing wrong? Hi, You are only finding the different combinations and not concerned about the ordering of the combinations. Suppose the females are f1, f2, f3 , f4 and f5, including f1 and f2 means you do not have to consider f1, f2 and f2, f1 as separate. You do it only once. Hence you have to use the combination formula.
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A team of 6 cooks is chosen from 8 men and 5 women. The team [#permalink]
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01 Oct 2013, 12:52
Hi,
I have the GMAT coming up on November 7th. I'm starting to panic because I get almost every PS problem that involves probabilities & selection wrong.
By selection I mean problems like this:
A team of 6 cooks is chosen from 8 men and 5 women. The team must have at least 2 men and at least 3 women. How many ways can this team be created?
A) 140 B) 320 C) 560 D) 700 E) 840
It's not that I don't get the formulas, I do. I just have no idea when to apply which formula. Or how you guys/ladies are looking at a problem and thinking "oh if I break this problem down into simpler terms, I can use formula ABC to solve it". It's exceedingly frustrating, because I'm getting about 90% of the problems incorrect. I just don't know what to do. I'm usually pretty good at math, I took the GRE about 7 years ago and was around 760 on the math section, but the GMAT questions just seem so different.
Last edited by Narenn on 01 Oct 2013, 21:24, edited 2 times in total.
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Re: No idea how to apply the PS formulae I'm learning here [#permalink]
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01 Oct 2013, 13:37
AccipiterQ wrote: Hi,
I have the GMAT coming up on November 7th. I'm starting to panic because I get almost every PS problem that involves probabilities & selection wrong.
By selection I mean problems like this:
A team of 6 cooks is chosen from 8 men and 5 women. The team must have at least 2 men and at least 3 women. How many ways can this team be created?
A) 140 B) 320 C) 560 D) 700 E) 840
It's not that I don't get the formulas, I do. I just have no idea when to apply which formula. Or how you guys/ladies are looking at a problem and thinking "oh if I break this problem down into simpler terms, I can use formula ABC to solve it". It's exceedingly frustrating, because I'm getting about 90% of the problems incorrect. I just don't know what to do. I'm usually pretty good at math, I took the GRE about 7 years ago and was around 760 on the math section, but the GMAT questions just seem so different. Question is not so difficult. Only thing matters is how do we simplify the information. We know the team must have ATLEAST 2 men and ATLEAST 3 women. So the possible combinations are 2(M) and 4(W) or 3(M) and 3(W) Case I : 2 Men and 4 Women > selection of 2 men from 8 men AND selection of 4 women from 5 women > 8 C 2 * 5 C 4 > 28*5 > 140 teams Case II : 3 Men and 3 Women > selection of 3 men from 8 men AND selection of 3 women from 5 women > 8 C 3 * 5 C 3 > 56*10 > 560 teams Total Number of teams = 140 + 560 = 700 > Choice D Hope that Helps! PS : For reference check my Permutations and Combinations article listed in my signature.
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Re: No idea how to apply the PS formulae I'm learning here [#permalink]
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01 Oct 2013, 13:39
Narenn wrote: AccipiterQ wrote: Hi,
I have the GMAT coming up on November 7th. I'm starting to panic because I get almost every PS problem that involves probabilities & selection wrong.
By selection I mean problems like this:
A team of 6 cooks is chosen from 8 men and 5 women. The team must have at least 2 men and at least 3 women. How many ways can this team be created?
A) 140 B) 320 C) 560 D) 700 E) 840
It's not that I don't get the formulas, I do. I just have no idea when to apply which formula. Or how you guys/ladies are looking at a problem and thinking "oh if I break this problem down into simpler terms, I can use formula ABC to solve it". It's exceedingly frustrating, because I'm getting about 90% of the problems incorrect. I just don't know what to do. I'm usually pretty good at math, I took the GRE about 7 years ago and was around 760 on the math section, but the GMAT questions just seem so different. Question is not so difficult. Only thing matters is how do we simplify the information. We know the team must have ATLEAST 2 men and ATLEAST 3 women. So the possible combinations are 2(M) and 4(W) or 3(M) and 3(W) Case I : 2 Men and 4 Women > selection of 2 men from 8 men AND selection of 4 women from 5 women > 8 C 2 * 5 C 4 > 28*5 > 140 teams Case II : 3 Men and 3 Women > selection of 3 men from 8 men AND selection of 3 women from 5 women > 8 C 3 * 5 C 3 > 56*10 > 560 teams Total Number of teams = 140 + 560 = 700 > Choice D Hope that Helps! Thanks for the reply; it's actually a question posted here elsewhere, I just was using it as a sample of the type of problem. The issue I have is that I have no idea how to apply these formulae or when to apply them.



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Re: No idea how to apply the PS formulae I'm learning here [#permalink]
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01 Oct 2013, 13:47
AccipiterQ wrote: Thanks for the reply; it's actually a question posted here elsewhere, I just was using it as a sample of the type of problem. The issue I have is that I have no idea how to apply these formulae or when to apply them. Ok. Refer the Article 'Permutations and Combinations' listed in my Signature below. I am sure it will help you understand when to apply which formula. Thanks Narenn
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Re: No idea how to apply the PS formulae I'm learning here [#permalink]
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01 Oct 2013, 17:51
Narenn wrote: AccipiterQ wrote: Thanks for the reply; it's actually a question posted here elsewhere, I just was using it as a sample of the type of problem. The issue I have is that I have no idea how to apply these formulae or when to apply them. Ok. Refer the Article 'Permutations and Combinations' listed in my Signature below. I am sure it will help you understand when to apply which formula. Thanks Narenn I just downloaded several of your PDFs, looks like I have some more study material! Thanks!



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Re: A team of 6 cooks is chosen from 8 men and 5 women. The team [#permalink]
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02 Oct 2013, 03:38
AccipiterQ wrote: Hi,
I have the GMAT coming up on November 7th. I'm starting to panic because I get almost every PS problem that involves probabilities & selection wrong.
By selection I mean problems like this:
A team of 6 cooks is chosen from 8 men and 5 women. The team must have at least 2 men and at least 3 women. How many ways can this team be created?
A) 140 B) 320 C) 560 D) 700 E) 840
It's not that I don't get the formulas, I do. I just have no idea when to apply which formula. Or how you guys/ladies are looking at a problem and thinking "oh if I break this problem down into simpler terms, I can use formula ABC to solve it". It's exceedingly frustrating, because I'm getting about 90% of the problems incorrect. I just don't know what to do. I'm usually pretty good at math, I took the GRE about 7 years ago and was around 760 on the math section, but the GMAT questions just seem so different. Merging similar topics. Please refer to the solutions above.
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Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
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