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I have trouble figuring out the part of when 2 men refuse to

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I have trouble figuring out the part of when 2 men refuse to [#permalink] New post 18 Oct 2005, 07:44
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I have trouble figuring out the part of when 2 men refuse to work together...can someone pls explain this in more detail...I can logically do it..but is there a formula for this kind of restriction

A group of 6 is chosen from 8 men and 5 women so as to contain at least 2 men and 3 women. How many different groups can be formed if two of the men refuse to server together?

a) 3510
b) 2620
c) 1404
d) 700
e) 635
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 [#permalink] New post 18 Oct 2005, 08:45
I am getting 910 combinations which is not in the answer list.
The possible combinations are:
2M 4W
3M 3W

if 2 men never work together, count 2 men as 1. total combinations are:
7C2 * 5C4 + 7C3 * 5C3 = 455
now 2 men(those counted as 1) can be picked in 2 ways.
total combinations: 455 * 2 = 910
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 [#permalink] New post 18 Oct 2005, 08:50
duttsit wrote:
I am getting 910 combinations which is not in the answer list.
The possible combinations are:
2M 4W
3M 3W

if 2 men never work together, count 2 men as 1. total combinations are:
7C2 * 5C4 + 7C3 * 5C3 = 455
now 2 men(those counted as 1) can be picked in 2 ways.
total combinations: 455 * 2 = 910


This is similar to the chair problems where two people cannot sit next to each other. However, I'm not sure what is the theory behind combining 2 men as 1 total combination. Can you explain the reasoning behind this method?

"count 2 men as 1. total combinations "
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 [#permalink] New post 18 Oct 2005, 09:16
well logicall speaking here is what I have done:

6 people to be choosen so
possible 8C3*5C3 + 8C2*5C3=700

well now I figured, well two men refuse to work together so there must be some combinations that need to be taken out of 700...

the only ans. choice less than 700 is E...
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 [#permalink] New post 18 Oct 2005, 09:21
ywilfred wrote:

This is similar to the chair problems where two people cannot sit next to each other. However, I'm not sure what is the theory behind combining 2 men as 1 total combination. Can you explain the reasoning behind this method?

"count 2 men as 1. total combinations "

Sorry. Guess my reasoning is wrong above. I will update once i find something.
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 [#permalink] New post 18 Oct 2005, 09:43
duttsit wrote:
ywilfred wrote:

This is similar to the chair problems where two people cannot sit next to each other. However, I'm not sure what is the theory behind combining 2 men as 1 total combination. Can you explain the reasoning behind this method?

"count 2 men as 1. total combinations "

Sorry. Guess my reasoning is wrong above. I will update once i find something.



Well, I solved it this way.

Two possible ways: 4W2M OR 3W3M
I'll call the two men who refuse to serve together John and Dick

Case 1: 4W2M. Can be broken into three subgroups.
Subgroup A: 2 men (excluding John and Dick) and 4 women
So we have 6C2*5C4 = 75 ways

Subgroup B: 2 men (John and one other) and 4 women
So we have 6C1*5C4 = 30

Subgroup C: 2 men (Dick and one other) and 4 women
SO we have another 30 ways

Case 2: 3M3W. This case can be further broken down in 3 subgroups.

Subgroup 1: 3 men (excluding John and Dick) and 3 women
So we have 6C3*5C3 = 200

SUbgroup 2: 3 men (John and 2 others) and 3 women
SO we have 6C2*5C3 = 150

Subgroup 3: 3 men (Dick and 2 others) and 3 women
So we have 6C2*5C3 = 150

Total = 635 ways

Of course, this method is rather long, but it's a logical way of thinking. But I'm game for any shortcuts. I'll probably only use this method if I don't have a short-cut, or I can't remember the short-cut during the test. :wink:
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 [#permalink] New post 18 Oct 2005, 09:57
I got 635


there are only combos of M and W

2Men 4Women
or
3Men 6Women


First calculate number of men that can be chosen given that two men can't work together

8C2=28 Now there is only 1 combination where workers say #1 and #2 can be together - subtract 1 you're left with 27 combinations

27*5C4=135

Now calculate number of 3 men that can be chosen from 8 given that workers #1 and #2 can't work together

8C3=56 We know that there are 6 combinations in which #1 and #2 are together:

1 3 2
1 4 2
1 5 2
1 6 2
1 7 2
1 8 2

Thus subtract 6 from 56 = 50

50*5C3 = 500

500+135=635
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 [#permalink] New post 18 Oct 2005, 10:08
ywilfred wrote:

Well, I solved it this way.

Two possible ways: 4W2M OR 3W3M
I'll call the two men who refuse to serve together John and Dick

Case 1: 4W2M. Can be broken into three subgroups.
Subgroup A: 2 men (excluding John and Dick) and 4 women
So we have 6C2*5C4 = 75 ways

Subgroup B: 2 men (John and one other) and 4 women
So we have 6C1*5C4 = 30

Subgroup C: 2 men (Dick and one other) and 4 women
SO we have another 30 ways

Case 2: 3M3W. This case can be further broken down in 3 subgroups.

Subgroup 1: 3 men (excluding John and Dick) and 3 women
So we have 6C3*5C3 = 200

SUbgroup 2: 3 men (John and 2 others) and 3 women
SO we have 6C2*5C3 = 150

Subgroup 3: 3 men (Dick and 2 others) and 3 women
So we have 6C2*5C3 = 150

Total = 635 ways

Of course, this method is rather long, but it's a logical way of thinking. But I'm game for any shortcuts. I'll probably only use this method if I don't have a short-cut, or I can't remember the short-cut during the test. :wink:


Good explanation. thanks.
I guess the number of steps can be reduced if we find the opposite first:

That is, find all combinations when two persons are ALWAYS together:

to chose 2M 4W, number of such arrangements will be:
6C0 * 5C4 = 5
{Number of combinations of n different things taken r at a time when p particular things always occur is n -pCr -p. }

similarly, to chose 3M 3W, number of arrangement when two men always together:
6C1 * 5C3 = 60

Total such combinations : 60 + 5 = 65
We need to subtract this from all arrangements (700 : see fresinha2 post below) to get required number:
700 - 65 = 635
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 [#permalink] New post 18 Oct 2005, 11:14
Titleist wrote:
Good explanation. thanks.
I guess the number of steps can be reduced if we find the opposite first:

That is, find all combinations when two persons are ALWAYS together:

to chose 2M 4W, number of such arrangements will be:
6C0 * 5C4 = 5
{Number of combinations of n different things taken r at a time when p particular things always occur is n -pCr -p. }

similarly, to chose 3M 3W, number of arrangement when two men always together:
6C1 * 5C3 = 60

Total such combinations : 60 + 5 = 65
We need to subtract this from all arrangements (700 : see fresinha2 post below) to get required number:
700 - 65 = 635


I like this approach! ![/quote]


Here is my confusion - Let us try to solve a simpler question
What are the combinations out of a group of 8 men, taken 3 at a time and two people always together.

so from your approach it is 8-2 C 3-2 => 6 C 1 = 6.

Now see it like this
ABCDEF XY are people X and Y always together. Conside X and Y one as Z.
Case 1 - when XY are not in the group selected Then 6 C 3 (3 people out of ABCDEF)
Case 2 when XY are in group then 6 C 1 ( 1 out of ABCDEF)

So the formula which you gave assumes that XY will always be in group.

I think less sleep is the root cause of all this ;)
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 [#permalink] New post 18 Oct 2005, 11:36
jainvineet wrote:
Titleist wrote:
Good explanation. thanks.
I guess the number of steps can be reduced if we find the opposite first:

That is, find all combinations when two persons are ALWAYS together:

to chose 2M 4W, number of such arrangements will be:
6C0 * 5C4 = 5
{Number of combinations of n different things taken r at a time when p particular things always occur is n -pCr -p. }

similarly, to chose 3M 3W, number of arrangement when two men always together:
6C1 * 5C3 = 60

Total such combinations : 60 + 5 = 65
We need to subtract this from all arrangements (700 : see fresinha2 post below) to get required number:
700 - 65 = 635


I like this approach! !



Here is my confusion - Let us try to solve a simpler question
What are the combinations out of a group of 8 men, taken 3 at a time and two people always together.

so from your approach it is 8-2 C 3-2 => 6 C 1 = 6.

Now see it like this
ABCDEF XY are people X and Y always together. Conside X and Y one as Z.
Case 1 - when XY are not in the group selected Then 6 C 3 (3 people out of ABCDEF)
Case 2 when XY are in group then 6 C 1 ( 1 out of ABCDEF)

So the formula which you gave assumes that XY will always be in group.

I think less sleep is the root cause of all this ;)[/quote]

Yes you are correct - that is why you must subtract the number in which people are Always together from the total possible number of combinations to get the number when the two people are never together.

If we take your question and change it to "What are the combinations out of a group of 8 men, taken 3 at a time and two people are never together."

It would be 8C3-6C1=50 possible combinations

Hope this clarifies Jain
  [#permalink] 18 Oct 2005, 11:36
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