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Combinatorics Question...(using "Thursday's w/ Ron"

by thp510 » Tue Feb 08, 2011 8:19 am
The gist of Ron's strategy from an old Thursday's With Ron Session on how to solve combinatorics problems 03DEC09 (https://www.manhattangmat.com/thursdays-with-ron.cfm):

1) Mark out the number of slots for each decision you have to make
2) Fill in each slot with the number of options
3) Multiply
*4) If order doesn't matter, take the number(s) of items where order doesn't matter and divide by it's factorial.

Question:
Greg, Marcia, Peter, Jan, Bobby, and Cindy go to a movie and sit next to each other in 6 adjacent seats in the front row of the theater. If Marcia and Jan will not sit next to each other, in how many different arrangements can the six people sit?

1) _ _ _ _ _ _ (six slots for the six seats)
2) 2 4 4 3 2 1 (The first "2" represents either Jan or Marcia, the first "4" represents Greg, Peter, Bobby and Cindy since they are the only ones that can sit next to either Jan or Marcia, the third "4" represents the four that are remaining after the first two sit down, etc, etc).
3) Multiply: 192
4) Does order matter here? Yes, this is an arrangement. So we stop. IMO: 192.

However, OA is 480. Why is my math wrong? I thought I followed Ron's method correctly? [/b]
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by GMATGuruNY » Tue Feb 08, 2011 8:39 am
thp510 wrote:The gist of Ron's strategy from an old Thursday's With Ron Session on how to solve combinatorics problems 03DEC09 (https://www.manhattangmat.com/thursdays-with-ron.cfm):

1) Mark out the number of slots for each decision you have to make
2) Fill in each slot with the number of options
3) Multiply
*4) If order doesn't matter, take the number(s) of items where order doesn't matter and divide by it's factorial.

Question:
Greg, Marcia, Peter, Jan, Bobby, and Cindy go to a movie and sit next to each other in 6 adjacent seats in the front row of the theater. If Marcia and Jan will not sit next to each other, in how many different arrangements can the six people sit?

1) _ _ _ _ _ _ (six slots for the six seats)
2) 2 4 4 3 2 1 (The first "2" represents either Jan or Marcia, the first "4" represents Greg, Peter, Bobby and Cindy since they are the only ones that can sit next to either Jan or Marcia, the third "4" represents the four that are remaining after the first two sit down, etc, etc).
3) Multiply: 192
4) Does order matter here? Yes, this is an arrangement. So we stop. IMO: 192.

However, OA is 480. Why is my math wrong? I thought I followed Ron's method correctly? [/b]
You're overly restricting the problem. For example, by placing 2 in the first position, you're saying that only J or M can sit there. But J and M can sit in any of the 6 seats; they just can't sit adjacent to each other.

Here's an efficient approach:

Good arrangements = total possible arrangements - bad arrangements.

Total arrangements = number of ways to arrange 6 distinct elements = 6! = 6*5*4*3*2*1 = 720.

Bad arrangements are those in which J and M sit next to each other. To determine the number of bad arrangements, we need to count the number of ways to arrange the following 5 elements: JM (as a unit) and the other 4 people. The number of ways to arrange 5 distinct elements = 5! = 5*4*3*2*1 = 120.

Since JM can be reversed to MJ, the result above needs to be doubled: 2*120 = 240.

Good arrangements = total possible arrangements - bad arrangements = 720-240 = 480.
Last edited by GMATGuruNY on Tue Feb 08, 2011 12:40 pm, edited 1 time in total.
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by Night reader » Tue Feb 08, 2011 12:37 pm
For sure the method of Mitch is the real brisk and effective, but just in case you want to walk through arrangements :)

six places (slots)

_/^\__/^\__/^\__/^\__/^\__/^\_ we want to know how to place Marcia and Jan not to be adjacent to each other

_/4\__/1\__/ 3 \__/3\__/ 2\__/1\_ ..72 ..----> now repeat this for three more times, when Marcia or Jan is placed in the middle

_/3\__/4\__/ 1 \__/3\__/ 2\__/1\_ ..72

_/3\__/2\__/ 4 \__/1\__/ 3\__/1\_ ..72

_/3\__/2\__/ 1 \__/4\__/ 1\__/3\_ ..72

and Marcia, Jan, either of them is on the ends

_/1\__/4\__/ 4 \__/3\__/ 2\__/1\_ ..96

_/4\__/3\__/ 2 \__/1\__/ 4\__/1\_ ..96

the total=72*4+96*2=288+192=480


GMATGuruNY wrote:
thp510 wrote:The gist of Ron's strategy from an old Thursday's With Ron Session on how to solve combinatorics problems 03DEC09 (https://www.manhattangmat.com/thursdays-with-ron.cfm):

1) Mark out the number of slots for each decision you have to make
2) Fill in each slot with the number of options
3) Multiply
*4) If order doesn't matter, take the number(s) of items where order doesn't matter and divide by it's factorial.

Question:
Greg, Marcia, Peter, Jan, Bobby, and Cindy go to a movie and sit next to each other in 6 adjacent seats in the front row of the theater. If Marcia and Jan will not sit next to each other, in how many different arrangements can the six people sit?

1) _ _ _ _ _ _ (six slots for the six seats)
2) 2 4 4 3 2 1 (The first "2" represents either Jan or Marcia, the first "4" represents Greg, Peter, Bobby and Cindy since they are the only ones that can sit next to either Jan or Marcia, the third "4" represents the four that are remaining after the first two sit down, etc, etc).
3) Multiply: 192
4) Does order matter here? Yes, this is an arrangement. So we stop. IMO: 192.

However, OA is 480. Why is my math wrong? I thought I followed Ron's method correctly? [/b]
You're overly restricting the problem. For example, by placing 2 in the first position, you're saying that only J or M can sit there. But J and M can sit in any of the 6 seats; they just can't sit adjacent to each other.

Here's an efficient approach:

Good arrangements = total possible arrangements - bad arrangements.

Total arrangements = number of ways to arrange 6 distinct elements = 6! = 6*5*4*3*2*1 = 720.

Bad arrangements are those in which J and M sit next to each other. To determine the number of bad arrangements, we need to count the number of ways to arrange 5 elements: JM (as a unit) and the other 4 people. Number of ways to arrange 5 distinct elements = 5! = 5*4*3*2*1 = 120.

Since JM can be reversed to MJ, the result above needs to be doubled: 2*120 = 240.

Good arrangements = total possible arrangements - bad arrangements = 720-240 = 480.
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by Jayanth2689 » Thu Jun 02, 2011 8:44 am
GMATGuruNY wrote:
thp510 wrote:The gist of Ron's strategy from an old Thursday's With Ron Session on how to solve combinatorics problems 03DEC09 (https://www.manhattangmat.com/thursdays-with-ron.cfm):

1) Mark out the number of slots for each decision you have to make
2) Fill in each slot with the number of options
3) Multiply
*4) If order doesn't matter, take the number(s) of items where order doesn't matter and divide by it's factorial.

Question:
Greg, Marcia, Peter, Jan, Bobby, and Cindy go to a movie and sit next to each other in 6 adjacent seats in the front row of the theater. If Marcia and Jan will not sit next to each other, in how many different arrangements can the six people sit?

1) _ _ _ _ _ _ (six slots for the six seats)
2) 2 4 4 3 2 1 (The first "2" represents either Jan or Marcia, the first "4" represents Greg, Peter, Bobby and Cindy since they are the only ones that can sit next to either Jan or Marcia, the third "4" represents the four that are remaining after the first two sit down, etc, etc).
3) Multiply: 192
4) Does order matter here? Yes, this is an arrangement. So we stop. IMO: 192.

However, OA is 480. Why is my math wrong? I thought I followed Ron's method correctly? [/b]
You're overly restricting the problem. For example, by placing 2 in the first position, you're saying that only J or M can sit there. But J and M can sit in any of the 6 seats; they just can't sit adjacent to each other.

Here's an efficient approach:

Good arrangements = total possible arrangements - bad arrangements.

Total arrangements = number of ways to arrange 6 distinct elements = 6! = 6*5*4*3*2*1 = 720.

Bad arrangements are those in which J and M sit next to each other. To determine the number of bad arrangements, we need to count the number of ways to arrange the following 5 elements: JM (as a unit) and the other 4 people. The number of ways to arrange 5 distinct elements = 5! = 5*4*3*2*1 = 120.

Since JM can be reversed to MJ, the result above needs to be doubled: 2*120 = 240.

Good arrangements = total possible arrangements - bad arrangements = 720-240 = 480.
Nice! In case of picking a group/team..and when order matters/does not matters, how should one handle the problem?
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by cans » Thu Jun 02, 2011 9:02 am
thp510 wrote: Question:
Greg, Marcia, Peter, Jan, Bobby, and Cindy go to a movie and sit next to each other in 6 adjacent seats in the front row of the theater. If Marcia and Jan will not sit next to each other, in how many different arrangements can the six people sit?
6 people. Total arrangements without any constraint = 6!
constraint - M and J will not sit next to each other. Suppose they sit together. Then we have 5 places and M and J can be arranged in 2! ways. Thus total of 5!*2!
Thus required answer = 6! - 5!*2 = 480
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by GMATGuruNY » Thu Jun 02, 2011 10:03 am
Jayanth2689 wrote:
GMATGuruNY wrote:
thp510 wrote:The gist of Ron's strategy from an old Thursday's With Ron Session on how to solve combinatorics problems 03DEC09 (https://www.manhattangmat.com/thursdays-with-ron.cfm):

1) Mark out the number of slots for each decision you have to make
2) Fill in each slot with the number of options
3) Multiply
*4) If order doesn't matter, take the number(s) of items where order doesn't matter and divide by it's factorial.

Question:
Greg, Marcia, Peter, Jan, Bobby, and Cindy go to a movie and sit next to each other in 6 adjacent seats in the front row of the theater. If Marcia and Jan will not sit next to each other, in how many different arrangements can the six people sit?

1) _ _ _ _ _ _ (six slots for the six seats)
2) 2 4 4 3 2 1 (The first "2" represents either Jan or Marcia, the first "4" represents Greg, Peter, Bobby and Cindy since they are the only ones that can sit next to either Jan or Marcia, the third "4" represents the four that are remaining after the first two sit down, etc, etc).
3) Multiply: 192
4) Does order matter here? Yes, this is an arrangement. So we stop. IMO: 192.

However, OA is 480. Why is my math wrong? I thought I followed Ron's method correctly? [/b]
You're overly restricting the problem. For example, by placing 2 in the first position, you're saying that only J or M can sit there. But J and M can sit in any of the 6 seats; they just can't sit adjacent to each other.

Here's an efficient approach:

Good arrangements = total possible arrangements - bad arrangements.

Total arrangements = number of ways to arrange 6 distinct elements = 6! = 6*5*4*3*2*1 = 720.

Bad arrangements are those in which J and M sit next to each other. To determine the number of bad arrangements, we need to count the number of ways to arrange the following 5 elements: JM (as a unit) and the other 4 people. The number of ways to arrange 5 distinct elements = 5! = 5*4*3*2*1 = 120.

Since JM can be reversed to MJ, the result above needs to be doubled: 2*120 = 240.

Good arrangements = total possible arrangements - bad arrangements = 720-240 = 480.
Nice! In case of picking a group/team..and when order matters/does not matters, how should one handle the problem?
There will always be fewer combinations possible than arrangements.

To illustrate:
There are 2 ways to arrange the letters A and B: AB and BA.
There is only 1 way to combine the letters A and B: AB.
BA represents the same combination.

When the order of the elements doesn't matter -- when we're choosing a team, for example -- we need to remove from our answer the duplicate combinations.
To remove the duplicate combinations, divide by (the number of elements in the combination)!:

Number of ways to arrange 4 elements from 6 choices = 6*5*4*3 = 720.
Number of ways to combine 4 elements from 6 choices = (6*5*4*3)/(4*3*2*1) = 15.

We divide by 4! = 4*3*2*1 because there are 4 elements in the combination.
Notice that the number of possible combinations is much less than the number of possible arrangements.
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by Jayanth2689 » Thu Jun 02, 2011 10:21 am
GMATGuruNY wrote:
Jayanth2689 wrote:
GMATGuruNY wrote:
thp510 wrote:The gist of Ron's strategy from an old Thursday's With Ron Session on how to solve combinatorics problems 03DEC09 (https://www.manhattangmat.com/thursdays-with-ron.cfm):

1) Mark out the number of slots for each decision you have to make
2) Fill in each slot with the number of options
3) Multiply
*4) If order doesn't matter, take the number(s) of items where order doesn't matter and divide by it's factorial.

Question:
Greg, Marcia, Peter, Jan, Bobby, and Cindy go to a movie and sit next to each other in 6 adjacent seats in the front row of the theater. If Marcia and Jan will not sit next to each other, in how many different arrangements can the six people sit?

1) _ _ _ _ _ _ (six slots for the six seats)
2) 2 4 4 3 2 1 (The first "2" represents either Jan or Marcia, the first "4" represents Greg, Peter, Bobby and Cindy since they are the only ones that can sit next to either Jan or Marcia, the third "4" represents the four that are remaining after the first two sit down, etc, etc).
3) Multiply: 192
4) Does order matter here? Yes, this is an arrangement. So we stop. IMO: 192.

However, OA is 480. Why is my math wrong? I thought I followed Ron's method correctly? [/b]
You're overly restricting the problem. For example, by placing 2 in the first position, you're saying that only J or M can sit there. But J and M can sit in any of the 6 seats; they just can't sit adjacent to each other.

Here's an efficient approach:

Good arrangements = total possible arrangements - bad arrangements.

Total arrangements = number of ways to arrange 6 distinct elements = 6! = 6*5*4*3*2*1 = 720.

Bad arrangements are those in which J and M sit next to each other. To determine the number of bad arrangements, we need to count the number of ways to arrange the following 5 elements: JM (as a unit) and the other 4 people. The number of ways to arrange 5 distinct elements = 5! = 5*4*3*2*1 = 120.

Since JM can be reversed to MJ, the result above needs to be doubled: 2*120 = 240.

Good arrangements = total possible arrangements - bad arrangements = 720-240 = 480.
Nice! In case of picking a group/team..and when order matters/does not matters, how should one handle the problem?
There will always be fewer combinations possible than arrangements.

To illustrate:
There are 2 ways to arrange the letters A and B: AB and BA.
There is only 1 way to combine the letters A and B: AB.
BA represents the same combination.

When the order of the elements doesn't matter -- when we're choosing a team, for example -- we need to remove from our answer the duplicate combinations.
To remove the duplicate combinations, divide by (the number of elements in the combination)!:

Number of ways to arrange 4 elements from 6 choices = 6*5*4*3 = 720.
Number of ways to combine 4 elements from 6 choices = (6*5*4*3)/(4*3*2*1) = 15.

We divide by 4! = 4*3*2*1 because there are 4 elements in the combination.
Notice that the number of possible combinations is much less than the number of possible arrangements.
Ok ! to pick from where you left..if i were to convert the above question to an anagram..

essentially the 4 elements are indistinguishable right? so..say i name each chosen element as Y and the non-chosen element as N..then the anagram would be

Y Y Y Y N N

hence, arranging YYYYNN = 6!/4!*2! (since order does not matter and to remove duplicates) = 15

also, this would be the permutation formula for unordered repetitive arrangements - nCk = n!/k!*(n-k)!

am i right in this regard? which method is faster to use??
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by GMATGuruNY » Thu Jun 02, 2011 10:37 am
Jayanth2689 wrote:
GMATGuruNY wrote:
Jayanth2689 wrote:
GMATGuruNY wrote:
thp510 wrote:The gist of Ron's strategy from an old Thursday's With Ron Session on how to solve combinatorics problems 03DEC09 (https://www.manhattangmat.com/thursdays-with-ron.cfm):

1) Mark out the number of slots for each decision you have to make
2) Fill in each slot with the number of options
3) Multiply
*4) If order doesn't matter, take the number(s) of items where order doesn't matter and divide by it's factorial.

Question:
Greg, Marcia, Peter, Jan, Bobby, and Cindy go to a movie and sit next to each other in 6 adjacent seats in the front row of the theater. If Marcia and Jan will not sit next to each other, in how many different arrangements can the six people sit?

1) _ _ _ _ _ _ (six slots for the six seats)
2) 2 4 4 3 2 1 (The first "2" represents either Jan or Marcia, the first "4" represents Greg, Peter, Bobby and Cindy since they are the only ones that can sit next to either Jan or Marcia, the third "4" represents the four that are remaining after the first two sit down, etc, etc).
3) Multiply: 192
4) Does order matter here? Yes, this is an arrangement. So we stop. IMO: 192.

However, OA is 480. Why is my math wrong? I thought I followed Ron's method correctly? [/b]
You're overly restricting the problem. For example, by placing 2 in the first position, you're saying that only J or M can sit there. But J and M can sit in any of the 6 seats; they just can't sit adjacent to each other.

Here's an efficient approach:

Good arrangements = total possible arrangements - bad arrangements.

Total arrangements = number of ways to arrange 6 distinct elements = 6! = 6*5*4*3*2*1 = 720.

Bad arrangements are those in which J and M sit next to each other. To determine the number of bad arrangements, we need to count the number of ways to arrange the following 5 elements: JM (as a unit) and the other 4 people. The number of ways to arrange 5 distinct elements = 5! = 5*4*3*2*1 = 120.

Since JM can be reversed to MJ, the result above needs to be doubled: 2*120 = 240.

Good arrangements = total possible arrangements - bad arrangements = 720-240 = 480.
Nice! In case of picking a group/team..and when order matters/does not matters, how should one handle the problem?
There will always be fewer combinations possible than arrangements.

To illustrate:
There are 2 ways to arrange the letters A and B: AB and BA.
There is only 1 way to combine the letters A and B: AB.
BA represents the same combination.

When the order of the elements doesn't matter -- when we're choosing a team, for example -- we need to remove from our answer the duplicate combinations.
To remove the duplicate combinations, divide by (the number of elements in the combination)!:

Number of ways to arrange 4 elements from 6 choices = 6*5*4*3 = 720.
Number of ways to combine 4 elements from 6 choices = (6*5*4*3)/(4*3*2*1) = 15.

We divide by 4! = 4*3*2*1 because there are 4 elements in the combination.
Notice that the number of possible combinations is much less than the number of possible arrangements.
Ok ! to pick from where you left..if i were to convert the above question to an anagram..

essentially the 4 elements are indistinguishable right? so..say i name each chosen element as Y and the non-chosen element as N..then the anagram would be

Y Y Y Y N N

hence, arranging YYYYNN = 6!/4!*2! (since order does not matter and to remove duplicates) = 15

also, this would be the permutation formula for unordered repetitive arrangements - nCk = n!/k!*(n-k)!

am i right in this regard? which method is faster to use??
I prefer the slot method:

1. Draw a slot for each position.
2. Determine the number of choices for each slot.
3. Multiply.

This method seems the most adaptable.
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by Jayanth2689 » Thu Jun 02, 2011 10:41 am
THANK YOU ! @GmatGuru
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by edvhou812 » Thu Jun 02, 2011 9:44 pm
cans wrote:
thp510 wrote: Question:
Greg, Marcia, Peter, Jan, Bobby, and Cindy go to a movie and sit next to each other in 6 adjacent seats in the front row of the theater. If Marcia and Jan will not sit next to each other, in how many different arrangements can the six people sit?
6 people. Total arrangements without any constraint = 6!
constraint - M and J will not sit next to each other. Suppose they sit together. Then we have 5 places and M and J can be arranged in 2! ways. Thus total of 5!*2!
Thus required answer = 6! - 5!*2 = 480
Dang, cans. Just take the test already. You freaking rock.
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by cans » Thu Jun 02, 2011 9:48 pm
edvhou812 wrote: Dang, cans. Just take the test already. You freaking rock.
Thanks but I am still working on my Verbal which needs to be improved a lot... :)
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by nafiul9090 » Fri Jun 03, 2011 10:03 am
hello guys,

nice discussion....combinatorics really killing me..

guys could you please shed some light on the following problem, though it was posted earlier, but i think the problem needs some expert opinions



Six mobsters have arrived at the theater for the premiere of the film "Goodbuddies." One of the mobsters, Frankie, is an informer, and he's afraid that another member of his crew, Joey, is on to him. Frankie, wanting to keep Joey in his sights, insists upon standing behind Joey in line at the concession stand, though not necessarily right behind him. How many ways can the six arrange themselves in line such that Frankie's requirement is satisfied?

A. 6

B. 24

C. 120

D. 360

E. 720

kinds regards

nafi

thanks in advance
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by Jayanth2689 » Fri Jun 03, 2011 9:25 pm
First, take the total number of ways of arranging 6 people without any constraints - 6! - 720 ways

The constraint - Joey has to stand behind Frankie irrespective of which position Frankie is in line. This means that Frankie can stand in the 1st, 2nd, 3rd, 4th and 5th positions only.

So if F is standing 1st, the rest of the them can arrange themselves in 5! ways - 120 ways.

Therefore, for all the 5 positions which F takes, there are 120 ways of arranging the others - 5*120 = 600 ways

Hence, the number of ways the 6 of them can arrange themselves is 720 - 600 = 120 ways (C)

Hope this is right.
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