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Need help with basic combinatorics problem

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Need help with basic combinatorics problem  [#permalink]

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New post 24 Sep 2018, 04:12
Hi all! I will be grateful if you could help me. First I saw a questions that says, if I have 4 Women and 3 Men and need to make a committee of 2 people, what's the probability the committee will have 2 women?

Could you confirm the three approaches below? Are they all right?

1. One can do (4/7)*(3/6) = 2/7
2. Also one can do #Have2WomenPermutations/#TotalPermutations and I think that's that's (4*3)/(7*6) = 2/7 - This is correct right?
3. Also one can do (#Total-#DoNotHave2WomenPermutations)/#TotalPermutations

I struggled with three, but I think I found it. Basically I think of Women and Men as letters WWWWMMM and I want to know how many ways there are to chose 2 that are not WW. What's the best way to find that? Is it by thinking - Pick2Male + PickWomanthenMale + PickMalethenWoman? So 3*2 +4*3 + 3*4? That's 6 + 12 + 12 = 30 So then (42-30)/42 = 2/7? Is that right?

Lastly, I had the thought below after finishing and I'd be grateful if anyone could provide clarity

Ok, so what if I want to know the possible combinations of 2 letters of a set WWWWMMM when order doesn't matter? Logically only 3 options are WM/MM/WW. I get it, but what is the formula that will give me that 3 I am seeking using combinations/permutations? This is driving me crazy! It reminds me to MISISSIPI but I am not able to apply similar strategies here.

I want to add that I think of "combinations" as "Permutations - Redundancies". So basically when I do 5C2 I think... there's 5*4 permutations and I need to eliminate 2! redundancies hence 10. Same with arranging a word such as IISSI we have 5! permutations and we can see the redundancies are going to be 3!2!. The inability to find a similar logic to the above problem is what's worrying me.

Thanks a lot.
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Re: Need help with basic combinatorics problem  [#permalink]

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New post 24 Sep 2018, 07:11
gmat800live wrote:
Hi all! I will be grateful if you could help me. First I saw a questions that says, if I have 4 Women and 3 Men and need to make a committee of 2 people, what's the probability the committee will have 2 women?

Could you confirm the three approaches below? Are they all right?

1. One can do (4/7)*(3/6) = 2/7
ok

2. Also one can do #Have2WomenPermutations/#TotalPermutations and I think that's that's (4*3)/(7*6) = 2/7 - This is correct right?
hmmmmm........ok- but I would rephrase it as #Have2WomenCombinations/#TotalCombinations
as in selecting a committee .... the order souldn't matter.
e.g., we have A,B,C,D
Q1. How many ways we can select 1 as president and another as VP ?
1 as P( this can be 4 people) = 4 & 2nd as VP(this can be 4-1=3)= 3 ways ==>total 4*3 ways or mathematically 4P2. Note : by mentioning P & VP we are imposing a restriction on order of selection . Thus its a permutation problem.
Q2. How many ways we can select 2 to form a games committee ?
we have (4*3)/2 as if we have selected AB; we also have BA there .... for each selection another one becomes redundant. Thus we divide by 2. Mathematically 4C2.[there is no difference in saying in the games committee we have April & Betty vs Betty & April]


3. Also one can do (#Total-#DoNotHave2WomenPermutations)/#TotalPermutations
Alright but - this would be looooonger metheod : DoNotHave2WomenPermutations)= 1- ( only men + possible men & women )

I struggled with three, but I think I found it. Basically I think of Women and Men as letters WWWWMMM and I want to know how many ways there are to chose 2 that are not WW. What's the best way to find that? Is it by thinking - Pick2Male + PickWomanthenMale + PickMalethenWoman? So 3*2 +4*3 + 3*4? That's 6 + 12 + 12 = 30 So then (42-30)/42 = 2/7? Is that right?

Lastly, I had the thought below after finishing and I'd be grateful if anyone could provide clarity

Ok, so what if I want to know the possible combinations of 2 letters of a set WWWWMMM when order doesn't matter? Logically only 3 options are WM/MM/WW. I get it, but what is the formula that will give me that 3 I am seeking using combinations/permutations? This is driving me crazy! It reminds me to MISISSIPI but I am not able to apply similar strategies here.

I want to add that I think of "combinations" as "Permutations - Redundancies". So basically when I do 5C2 I think... there's 5*4 permutations and I need to eliminate 2! redundancies hence 10. Same with arranging a word such as IISSI we have 5! permutations and we can see the redundancies are going to be 3!2!. The inability to find a similar logic to the above problem is what's worrying me.

How about I rephrase ur 2Qs and let you answer :
Q1. We have a set of 7 letters (WWWWMMM). In how many different ways we can arrange them ?
Q2. Ivanisevic- Family has 3 members & Saunders-Family has 2. In how many ways we can form a committee of 2. ( consider all Ivanisevic(s) & Saunders(s) are identical selection wrt their family0- i.e., if I select I1 , that selection must not be considered as a different selection than I2 or I3. Sililarly S1==S2. But I !=S)

Hope it helps.


Thanks a lot.




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Re: Need help with basic combinatorics problem  [#permalink]

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New post 25 Sep 2018, 22:37
Hi gmat800live

Aside from the great explanation provided by u1983

I highly recommend you to watch the Thursday's with Ron video on Combinatorics.

It seems to me your understanding is lacking in differentiating between "Order matters" & "Order doesn't matter" and hence when to use permutation or combination. Ron Purewal explains these concepts in the most basic manner.

Its a great video lesson, that will enable you to solve combinatorics problems of all level with ease.

Video link:

P.s The video starts abruptly, so don't let that bother you.


Thanks,
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Need help with basic combinatorics problem  [#permalink]

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New post 27 Sep 2018, 10:54
gmat800live wrote:
Hi all! I will be grateful if you could help me. First I saw a questions that says, if I have 4 Women and 3 Men and need to make a committee of 2 people, what's the probability the committee will have 2 women?

Could you confirm the three approaches below? Are they all right?

1. One can do (4/7)*(3/6) = 2/7

Correct!

Quote:
2. Also one can do #Have2WomenPermutations/#TotalPermutations and I think that's that's (4*3)/(7*6) = 2/7 - This is correct right?

This is slightly more complicated. You got the right answer, but not quite for the right reason.

4*3 is what you would do if the order did matter. For instance, suppose your four women are named Anne, Beth, Cassidy, and Dora. Here are your 4x3 options:

Anne and Beth
Anne and Cassidy
Anne and Dora
Beth and Anne
Beth and Cassidy
Beth and Dora
Cassidy and Anne
Cassidy and Beth
Cassidy and Dora
Dora and Anne
Dora and Beth
Dora and Cassidy

See how that's a little weird? We counted the 'Anne and Beth' pair twice, we counted the 'Beth and Dora' pair twice, etc. This is what happens when you act as if order matters, but it actually doesn't.

Technically, the number of pairs would be (4 total women)! / (2 women in the pair)!(2 women not in the pair)!

Which reduces to (4x3x2x1)/(2x1x2x1) = 6 pairs.

Similarly, the total number of pairs is (7 total people)!/(2 people in the pair)!(5 people not in the pair)! or 21 pairs.

6/21 = 2/7, which is the same as the answer you got. The lesson is that when you're doing a problem like this, technically, it doesn't matter how you deal with order, as long as you do it the same way in both the numerator and the denominator. But if the problem just asked you how many committees of two women there could be, the answer would be 6, not 12.

Quote:
3. Also one can do (#Total-#DoNotHave2WomenPermutations)/#TotalPermutations


Finally, this is definitely not the right way to do this problem! You would technically get the right answer, but you saw how much more complicated it is. This strategy is only efficient when it's easier to count the 'bad' cases than the 'good' cases. In this problem, it's much easier to count the pairs of women than to count 'all of the other different pairs,' as you saw.

Here's a different question, though, where this strategy would be better:

"How many different committees could be made that include at least one woman?"

In this case, the 'good' cases are cases with either one or two women, but the 'bad' cases are the ones with two men. It's easier to count the pairs of men, so it would be easier, in this case, to count the male-only pairs and subtract.

Quote:
Lastly, I had the thought below after finishing and I'd be grateful if anyone could provide clarity

Ok, so what if I want to know the possible combinations of 2 letters of a set WWWWMMM when order doesn't matter? Logically only 3 options are WM/MM/WW. I get it, but what is the formula that will give me that 3 I am seeking using combinations/permutations? This is driving me crazy! It reminds me to MISISSIPI but I am not able to apply similar strategies here.


Here's how the MISSISSIPPI problem works:

"How many ways can all of the letters of the word Mississippi be rearranged?"

(11 letters)!/(1 M)!(4 S)!(4 I)!(2 P)! = 11!/(1!x4!x4!x2!) = 11x10x9x8x7x6x5x4x3x2x1 / (1 x 4x3x2x1 x 4x3x2x1 x 2x1)

= 11x10x9x8x7x6x5x4x3x2x1 / (1 x 4x3x2x1 x 4x3x2x1 x 2x1)

= 11x10x9x8x7x6x5 / (4x3x2 x 2)

= 11x10x9x8x7x6x5 / (4x3x2 x 2)

= 11x10x9x8x7x5 / (4 x 2)

= 11x10x9x8x7x5 / (4 x 2)

= 11x10x9x7x5 = 34,650

Your question isn't really the same as this. You're not trying to rearrange all seven letters of WWWWMMM. Instead, you're asking a slightly different question: "How many different two-letter words can be made from the letters WWWWMMM"?

As it turns out, this makes the question a lot more complex! I'm not going to go through the math, because the only situation where the GMAT would ask a question like this, would be if you could count all of the possibilities by hand (which is what you did when you listed WW, WM, MW.) There isn't a single, straightforward equation that'll solve every problem like this. So, I wouldn't worry too much about it.




Quote:
I want to add that I think of "combinations" as "Permutations - Redundancies". So basically when I do 5C2 I think... there's 5*4 permutations and I need to eliminate 2! redundancies hence 10. Same with arranging a word such as IISSI we have 5! permutations and we can see the redundancies are going to be 3!2!. The inability to find a similar logic to the above problem is what's worrying me.


This is exactly correct. (Well, you should be thinking 'permutations divided by redundancies', not subtraction, but your example shows that you're understanding it correctly.)
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Need help with basic combinatorics problem &nbs [#permalink] 27 Sep 2018, 10:54
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