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# PS Combinatorics

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Joined: 06 Apr 2013
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PS Combinatorics [#permalink]  24 Apr 2013, 13:11
Hi Guys,

I am new to this community and my GMAT will take place in 12 days.
I have question concerning two diffierent PS Combinatorics Questions. For my they seem to be identical dispite the round table and the number of items.
But they are solved in two different ways.

Here are the two Problems:

1. In how many ways can 11 # signs and 8* signs be arranged in a row so that no two * signs come together?

Answer Should be: (taken for the GMAT Club Combinatorics Set)

1st we place the 11#, now there are 10 places between them and 2 on the extreme left and extreme right of them, total places = 10 + 2 = 12
If * is placed at any of these 12 places, no 2 *'s will be together
so the number ways will be 12C8 = 12C4 = 495

2. Seven men and five women have to sit around a circular table so that no 2 women are together. In how many different ways can this be done?

Answer form Bunnel (which I totally understand):

# of arrangements of 7 men around a table is (7-1)!=6!;
There will be 7 possible places for women between them, 7 empty slots. # of ways to choose in which 5 slots women will be placed is C^5_7=21;
# of arrangements of 5 women in these slots is 5!;

So total: 6!*21*5!=1,814,400.

In my point of view the first problem should be solved with the same approach. Like this:

11! * 8! * 11C8 = very very big number....

I am a little confused. May somebody can help me? I also have problems to understand when we consider the objects like the # sings to be all different e.g #1 ; #2 etc. and when they all are just # so #=# for every #.

I hope you can understand my question.

Kind regards,
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Re: PS Combinatorics [#permalink]  24 Apr 2013, 21:04
1
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Expert's post
Cityboy wrote:
Hi Guys,

I am new to this community and my GMAT will take place in 12 days.
I have question concerning two diffierent PS Combinatorics Questions. For my they seem to be identical dispite the round table and the number of items.
But they are solved in two different ways.

Here are the two Problems:

1. In how many ways can 11 # signs and 8* signs be arranged in a row so that no two * signs come together?

Answer Should be: (taken for the GMAT Club Combinatorics Set)

1st we place the 11#, now there are 10 places between them and 2 on the extreme left and extreme right of them, total places = 10 + 2 = 12
If * is placed at any of these 12 places, no 2 *'s will be together
so the number ways will be 12C8 = 12C4 = 495

2. Seven men and five women have to sit around a circular table so that no 2 women are together. In how many different ways can this be done?

Answer form Bunnel (which I totally understand):

# of arrangements of 7 men around a table is (7-1)!=6!;
There will be 7 possible places for women between them, 7 empty slots. # of ways to choose in which 5 slots women will be placed is C^5_7=21;
# of arrangements of 5 women in these slots is 5!;

So total: 6!*21*5!=1,814,400.

In my point of view the first problem should be solved with the same approach. Like this:

11! * 8! * 11C8 = very very big number....

I am a little confused. May somebody can help me? I also have problems to understand when we consider the objects like the # sings to be all different e.g #1 ; #2 etc. and when they all are just # so #=# for every #.

I hope you can understand my question.

Kind regards,

Hey There,

The key thing to be noted here is, In problem one the objects to be arranged are identical; 11# signs and 8 * signs are identical in their respective groups. e.g. If you want to arrange 11 # signs in a row, you can do so in only one way because they are identical to each other. You can not distinguish a particular # sign from rest of the signs, whereas in the second problem, objects to be arranged are all different, so in this case you will get many number of ways to arrange those different objects. e.g. 4 persons you can arrange in 4! ways. This is the reason why answer of the 2nd ques is much more greater than that of the 1st

back to the Ques 1 :- In how many ways can 11 # signs and 8* signs be arranged in a row so that no two * signs come together?

jobs to be done
1) Arrangement of 11 # signs in a row ---------> You have only one option to arrange those signs, since they are identical
Having arranged them, now we have 12 places where we can place the 8 * signs
2) Selection of 8 places from 12 places to place the 8 * signs ------------> you can do this in 12C8 ways i.e. 495 ways
3) Arrangement of 8 * signs at 8 places selected earlier ---------> You have only one option to arrange those signs, since they are identical

Total Number of Arrangement{By Fundamental Principle of multiplication} = 1 X 495 X 1 = 495

Ques 2 :- 7 men and 5 women have to sit around a circular table so that no 2 women are together. In how many different ways can this be done?

jobs to be done
1) arrangement of 7 men around a table ----------> This you can do in (7-1)! = 6! ways. Because now you are doing circular arrangement and unlike linear arrangements, circulation arrangements dont have beginning or end.
Having arranged them, you have 7 places where you can arrange 5 women
2) selection of 5 places from 7 places where you can arrange women ----------> This can be done in 7C5 ways
3) Arrangement of 5 women at 5 places selected earlier-------> 5! ways

Total Number of Arrangement{By Fundamental Principle of multiplication} = 6! X 7C5 X 5! = 1814400

Regards,

Narenn
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Intern
Joined: 06 Apr 2013
Posts: 3
Location: United States
Concentration: Finance, General Management
GMAT Date: 05-07-2013
GPA: 3.65
Followers: 0

Kudos [?]: 5 [0], given: 3

Re: PS Combinatorics [#permalink]  25 Apr 2013, 02:30
First of all: Thanks a lot for this amazing answer.

But I still have big touble with the Combinatoric Tasks:

For Example:

There are 12 people in a room: 4 couples and 4 singles. picking 4 people at random, which is the probability of picking only one couple?

I understand that the Total amount of combinations of a Group of four is 12C4= 33*15
I also the the point that I can get 4 different couples = 4
But I am confused how we calculate the whole denomerator:

We have three option:

1. Combining one pair with 2 Singles (so we take 2 out of 4 which is 4C2 = 6 or not?)
2. Combining one pair with one guy of a another pair and one single (so we take one out of four singles 4C1 and one person form the remaining 6 who are in a couple which is 6C1 = 6) Both together is 4*6 = 24
3. Combining one pair with one guy of a pair and one guy of another pair ( we take one out of the 6 people who live in a couple which is 6C1= 6 and then again one out of the 4 remaining people who are a couple which is 4C1= 4) leads to 24

In the end I got 4*(24+24+6)= 216

216/33*11 = Does not lead to the official answer. OA seems to be 56/165 (taken form this discussion in the forum unfortuntaley I can not posts links...)

I really do not understand how to build up the ways of chosing one single and one guy of a couple and so on. I have this porblem in general. Can someone give me an advice how to think about such kind of problems?

Are there more such problems available?

Kind regards

Last edited by Cityboy on 25 Apr 2013, 04:52, edited 1 time in total.
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Re: PS Combinatorics [#permalink]  25 Apr 2013, 04:48
Expert's post
There are 12 people in a room: 4 couples and 4 singles. picking 4 people at random, which is the probability of picking only one couple?

Probability = $$\frac{Desired outcomes}{Total Outcomes}$$
Total Outcomes = selection of 4 persons among 12 persons = 12C4 = 495

Desired Outcomes = Selection of 1 Couple AND selection of 2 persons from remaining 10 persons (Condition - The two selected should not be a couple)

1 couple can be selected from 4 couples in 4C1 ways.

Now we are left with 3 couples and 4 singles (Total 10 Persons) from which we have to select 2 persons. (Condition - The two selected should not be a couple)

This can be calculated as = Total number of ways selecting 2 persons from 10 persons - Number of ways selecting a couple from available couples
= 10C2 - 3C2 = 42

Lets see how this arrived

Let these 10 persons be m1, f1, m2, f2, m3, f3, s1, s2, s3, s4

If you chose m1, you can choose any one of the remaining 9 persons except for f1--------> So the available options are 8
If you chose f1, you can choose any one of the remaining 9 persons except for m1--------> So the available options are 8
If you chose m2, you can choose any one of the remaining 9 persons except for f2, m1, f1 --------> So the available options are 6
If you chose f2, you can choose any one of the remaining 9 persons except for m2, m1, f1 --------> So the available options are 6
If you chose m3, you can choose any one of the remaining 9 persons except for f3, m1, m2, f1, f2 --------> So the available options are 4
If you chose f3, you can choose any one of the remaining 9 persons except for m3, m1, m2, f1, f2 --------> So the available options are 4
If you chose s1, you can choose any one of the remaining 9 persons except for m1, m2, m3, f1, f2, f3, f4 --------> So the available options are 3
If you chose s2, you can choose any one of the remaining 9 persons except for m1, m2, m3, f1, f2, f3, f4, s1--------> So the available option is 2
If you chose s3, you can choose any one of the remaining 9 persons except for m1, m2, m3, f1, f2, f3, f4, s1, s2--------> So the available options are 1
If you chose s3, you can choose any one of the remaining 9 persons except for m1, m2, m3, f1, f2, f3, f4, s1, s2, s3--------> So the available options are 0

Total 8+8+6+6+4+4+3+3+3 = 42 options

Total Desired outcomes [Fundamental Principle of Multiplication] = 4C1 X 42 = 168

Probablity = $$\frac{168}{495}$$

What is the OA ?

Narenn
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Last edited by Narenn on 25 Apr 2013, 05:18, edited 1 time in total.
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Re: PS Combinatorics [#permalink]  25 Apr 2013, 05:12
HI I Edited my post:

Unfortunately I cannot post links. You can find it nif you google "here are 12 people in a room: 4 couples and 4 singles. picking 4 people at random, which is the probability of picking only one couple?"

Big thanks again for your explanation!!!

So I have ten people remaining and first of all I can choose m1 or f1 so I still can combine m1 or f1 with 8 other people. (therefore 8 times 2)
In the next step I assume that I have choosen f1 or m1 before and then chose f2 or m2 and thus I can combine each of them with 6 other people? Then same for m3 and f3. Therefore 6 times 2
So right now I have already choosen m1, m2, m3 or f1,f2,f3 thus have s1,s2,s3,s4 left. So I can combine s1 with either s2,s3,s4 (or s2 with s1,s4,s3 or s3 with s2,s1,s4) which leads to 3x3 options.

Am I right?
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Re: PS Combinatorics [#permalink]  25 Apr 2013, 05:22
Expert's post
Cityboy wrote:
HI I Edited my post:

Unfortunately I cannot post links. You can find it nif you google "here are 12 people in a room: 4 couples and 4 singles. picking 4 people at random, which is the probability of picking only one couple?"

Big thanks again for your explanation!!!

So I have ten people remaining and first of all I can choose m1 or f1 so I still can combine m1 or f1 with 8 other people. (therefore 8 times 2)
In the next step I assume that I have choosen f1 or m1 before and then chose f2 or m2 and thus I can combine each of them with 6 other people? Then same for m3 and f3. Therefore 6 times 2
So right now I have already choosen m1, m2, m3 or f1,f2,f3 thus have s1,s2,s3,s4 left. So I can combine s1 with either s2,s3,s4 (or s2 with s1,s4,s3 or s3 with s2,s1,s4) which leads to 3x3 options.

Am I right?

It should be 168/495 or 56/165 and not earlier that i posted. There was an error in the calculation. I have just corrected it.

Regards,

Narenn
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Re: PS Combinatorics   [#permalink] 25 Apr 2013, 05:22
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