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sudhir3127
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P&C !!!1
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Source: Beat The GMAT — Problem Solving |
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sudhir3127
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here are the Answer choicessudhir3127 wrote:A boats crew consists of 8 men . 3 of whom can only row on one side and 2 only on the other, Find the number of ways in which the crew can be arranged...
Answers after few discussions.
a.8!/6!*5!
b.8!/2!3!
c.3(4!)*(4!)
d.4!/3!*2!
I don't think the answer is exactly any of those listed. I think if B had 2!3!3! in the denominator than it would be correct.
On one side you have 3! ways the 3 men can be arranged and on the other side you have 2! ways the 2 men can be arranged. You have 3 more places that can be arranged 3! ways.
So, I think the answer is
8!/2!3!3! = 60 ways.
On one side you have 3! ways the 3 men can be arranged and on the other side you have 2! ways the 2 men can be arranged. You have 3 more places that can be arranged 3! ways.
So, I think the answer is
8!/2!3!3! = 60 ways.
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anju
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What is the 5th option?sudhir3127 wrote:here are the Answer choicessudhir3127 wrote:A boats crew consists of 8 men . 3 of whom can only row on one side and 2 only on the other, Find the number of ways in which the crew can be arranged...
Answers after few discussions.
a.8!/6!*5!
b.8!/2!3!
c.3(4!)*(4!)
d.4!/3!*2!
My answer is 9 * 4! * 4!
Explanation:
There are 8 men say A B C D E F G H
On 1 side 2 men has to be seated along with 2 other (from the pool of 3 men who can row both sides)
Let A & B be 2 men who can row on 1 side
Let F, G & H be 3 men who can row on the second side
and let C, D n E be 3 men who can row both sides
so the anagram is for 1 side is (with 2 men on one side)
1 2 3 4
A B C D - 4! ways can be seated
A B C E - 4! ways can be seated
A B D E - 4! ways can be seated
so for 1 side 3*4!
Let's look at the anagram for the second side:
so the anagram is for 2 side is (with 3 men on one side)
5 6 7 8
F G H E - 4! ways can be seated
F G H D - 4! ways can be seated
F G H C - 4! ways can be seated
so for 2 side 3*4!
In all there are 1 side * 2 side arrangements so 3 * 4! * 3 * 4! = 9 * 4! * 4!
I am not sure if this approach is correct. What's the OA?
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gabriel
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I agree with this answer. But why is it that we are assuming that there should be 4 people on each side, the question says nothing of that sort. I have a feeling that the question has not been worded properly.LSB wrote:I would say the answer is C
You can arrange each side using 4!
But you have 3 guys that are flexible and can sub anywhere.
So 3*4!*4!
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gabriel
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Anju there is a mistake in your solution. Some of the combinations wont work and hence you are over counting. For example the combination of ABCD (The first combination in the first category) and FGHC (The last combination in the second category) is not possible because C cant be on both the sides.anju wrote:What is the 5th option?sudhir3127 wrote:here are the Answer choicessudhir3127 wrote:A boats crew consists of 8 men . 3 of whom can only row on one side and 2 only on the other, Find the number of ways in which the crew can be arranged...
Answers after few discussions.
a.8!/6!*5!
b.8!/2!3!
c.3(4!)*(4!)
d.4!/3!*2!
My answer is 9 * 4! * 4!
Explanation:
There are 8 men say A B C D E F G H
On 1 side 2 men has to be seated along with 2 other (from the pool of 3 men who can row both sides)
Let A & B be 2 men who can row on 1 side
Let F, G & H be 3 men who can row on the second side
and let C, D n E be 3 men who can row both sides
so the anagram is for 1 side is (with 2 men on one side)
1 2 3 4
A B C D - 4! ways can be seated
A B C E - 4! ways can be seated
A B D E - 4! ways can be seated
so for 1 side 3*4!
Let's look at the anagram for the second side:
so the anagram is for 2 side is (with 3 men on one side)
5 6 7 8
F G H E - 4! ways can be seated
F G H D - 4! ways can be seated
F G H C - 4! ways can be seated
so for 2 side 3*4!
In all there are 1 side * 2 side arrangements so 3 * 4! * 3 * 4! = 9 * 4! * 4!
I am not sure if this approach is correct. What's the OA?
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anju
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Thanks for pointing it out gabriel. After some thoughts, I guess it should begabriel wrote:Anju there is a mistake in your solution. Some of the combinations wont work and hence you are over counting. For example the combination of ABCD (The first combination in the first category) and FGHC (The last combination in the second category) is not possible because C cant be on both the sides.anju wrote:What is the 5th option?sudhir3127 wrote:here are the Answer choicessudhir3127 wrote:A boats crew consists of 8 men . 3 of whom can only row on one side and 2 only on the other, Find the number of ways in which the crew can be arranged...
Answers after few discussions.
a.8!/6!*5!
b.8!/2!3!
c.3(4!)*(4!)
d.4!/3!*2!
My answer is 9 * 4! * 4!
Explanation:
There are 8 men say A B C D E F G H
On 1 side 2 men has to be seated along with 2 other (from the pool of 3 men who can row both sides)
Let A & B be 2 men who can row on 1 side
Let F, G & H be 3 men who can row on the second side
and let C, D n E be 3 men who can row both sides
so the anagram is for 1 side is (with 2 men on one side)
1 2 3 4
A B C D - 4! ways can be seated
A B C E - 4! ways can be seated
A B D E - 4! ways can be seated
so for 1 side 3*4!
Let's look at the anagram for the second side:
so the anagram is for 2 side is (with 3 men on one side)
5 6 7 8
F G H E - 4! ways can be seated
F G H D - 4! ways can be seated
F G H C - 4! ways can be seated
so for 2 side 3*4!
In all there are 1 side * 2 side arrangements so 3 * 4! * 3 * 4! = 9 * 4! * 4!
I am not sure if this approach is correct. What's the OA?
4!*4! which is ABCD combination * FGHE combination +
4!*4! which is ABCE combination * FGHG combination +
4!*4! which is ABDE combination * FGHC combination which is equivalent to
3*4!*4!
I am sure there would be a better way to solve this question? Any thoughts?
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sudhir3127
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the OA is c.sudhir3127 wrote:A boats crew consists of 8 men . 3 of whom can only row on one side and 2 only on the other, Find the number of ways in which the crew can be arranged...
Answers after few discussions.
LSB has done in the best and shortest way .
ur right anju... its 3*4!*4!
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4meonly
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agreegabriel wrote: I have a feeling that the question has not been worded properly.
I began to count all possibilities - when there are 1 man on each side, 2 men on each side, so on...
