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Combinations

by akash singhal » Thu Nov 12, 2015 9:12 pm
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. How many ways
can the six arrange themselves in line such that Frankie's requirement is satisfied?


6
24
120
360
720

OE D

Please explain how and if possible through two processes

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by Brent@GMATPrepNow » Thu Nov 12, 2015 10:24 pm
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
If we remove the restriction that Frankie must stand behind Joey, then the mobsters can be arranged in 6! (720) different ways.
Now, of these 720 different arrangements, it is clear that Frankie will be standing behind Joey half the time, and Joey will be standing behind Frankie half the time.

So, the answer is 720/2 or 360.

Answer = D

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by Brent@GMATPrepNow » Thu Nov 12, 2015 10:25 pm
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by gmatbeater1989 » Mon Nov 23, 2015 11:27 am
I tried using the slot method here but didn't get very far. Can you use it with this question?

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by GMATGuruNY » Mon Nov 23, 2015 11:44 am
An alternate approach:

Let the 6 mobsters be A, B, C, D, F and J.
Direction of the line:
Front....Back.

Number of options for A = 6. (Any of the 6 positions.)
Number of options for B = 5. (Any of the 5 remaining positions.)
Number of options for C = 4. (Any of the 4 remaining positions.)
Number of options for D = 3. (Any of the 3 remaining positions.)

Of the two remaining positions, F must occupy the one more to the right, so that he can keep an eye on J.
Number of options for F = 1. (Of the two remaining positions, the one more to the right.)
Number of options for J = 1. (One position left.)

To combine these options, we multiply:
6*5*4*3*1*1 = 360.

The correct answer is D.
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by GMATGuruNY » Mon Nov 23, 2015 12:12 pm
gmatbeater1989 wrote:I tried using the slot method here but didn't get very far. Can you use it with this question?
Let the six positions be as follows:
BACK _ _ _ _ _ _ FRONT
Since F must stand behind J, J must be positioned to the right of F.

Case 1: F _ _ _ _ _
Since the 5 remaining slots may be occupied by any arrangement of the 5 remaining mobsters, the slots look as follows:
F * 5 * 4 * 3 * 2 * 1.
Multiplying the values in red, we get;
5*4*3*2*1 = 120.

Case 2: _ F _ _ _ _
Number of options for the first slot = 4. (Of the 5 remaining mobsters, anyone but J.)
Since the 4 remaining slots may be occupied by any arrangement of the 4 remaining mobsters, the slots look as follows:
4 * F * 4 * 3 * 2 * 1.
Multiplying the values in red, we get;
4*4*3*2*1 = 96.

Case 3: _ _ F _ _ _
Number of options for the first slot = 4. (Of the 5 remaining mobsters, anyone but J.)
Number of options for the second slot = 3. (Of the 4 remaining mobsters, anyone but J.)
Since the 3 remaining slots may be occupied by any arrangement of the 3 remaining mobsters, the slots look as follows:
4 * 3 * F * 3 * 2 * 1.
Multiplying the values in red, we get;
4*3*3*2*1 = 72.

Case 4: _ _ _ F _ _
Number of options for the first slot = 4. (Of the 5 remaining mobsters, anyone but J.)
Number of options for the second slot = 3. (Of the 4 remaining mobsters, anyone but J.)
Number of options for the third slot = 2. (Of the 3 remaining mobsters, anyone but J.)
Since the 2 remaining slots may be occupied by any arrangement of the 2 remaining mobsters, the slots look as follows:
4 * 3 * 2 * F * 2 * 1.
Multiplying the values in red, we get;
4*3*2*2*1 = 48.

Case 5: _ _ _ _ F J
Since the 4 remaining slots may be occupied by any arrangement of the 4 remaining mobsters, the slots look as follows:
4 * 3 * 2 * 1 * F * J.
Multiplying the values in red, we get;
4*3*2*1 = 24.

Total ways = 120 + 96 + 72 + 48 + 24 = 360.

The methods offered in the posts above seem faster.
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by gmatbeater1989 » Tue Nov 24, 2015 10:36 am
GMATGuruNY wrote:An alternate approach:

Let the 6 mobsters be A, B, C, D, F and J.
Direction of the line:
Front....Back.

Number of options for A = 6. (Any of the 6 positions.)
Number of options for B = 5. (Any of the 5 remaining positions.)
Number of options for C = 4. (Any of the 4 remaining positions.)
Number of options for D = 3. (Any of the 3 remaining positions.)

Of the two remaining positions, F must occupy the one more to the right, so that he can keep an eye on J.
Number of options for F = 1. (Of the two remaining positions, the one more to the right.)
Number of options for J = 1. (One position left.)

To combine these options, we multiply:
6*5*4*3*1*1 = 360.

The correct answer is D.
nice!

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by Matt@VeritasPrep » Fri Nov 27, 2015 2:13 am
gmatbeater1989 wrote:
GMATGuruNY wrote:An alternate approach:

Let the 6 mobsters be A, B, C, D, F and J.
Direction of the line:
Front....Back.

Number of options for A = 6. (Any of the 6 positions.)
Number of options for B = 5. (Any of the 5 remaining positions.)
Number of options for C = 4. (Any of the 4 remaining positions.)
Number of options for D = 3. (Any of the 3 remaining positions.)

Of the two remaining positions, F must occupy the one more to the right, so that he can keep an eye on J.
Number of options for F = 1. (Of the two remaining positions, the one more to the right.)
Number of options for J = 1. (One position left.)

To combine these options, we multiply:
6*5*4*3*1*1 = 360.

The correct answer is D.
nice!
We really want to avoid such an approach if we can, though: the slots method exists to make slot-friendly problems easier, not to make slot-hostile problems more time-consuming!

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by GMATsid2016 » Wed Nov 23, 2016 9:45 am
Now, of these 720 different arrangements, it is clear that Frankie will be standing behind Joey half the time, and Joey will be standing behind Frankie half the time.

So, the answer is 720/2 or 360.

Answer = D

Hi Brent ,

Everything is clear, but can you please explain more above part?

why 720/2?

Thanks,

Sid

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by Brent@GMATPrepNow » Wed Nov 23, 2016 11:19 am
GMATsid2016 wrote:
Now, of these 720 different arrangements, it is clear that Frankie will be standing behind Joey half the time, and Joey will be standing behind Frankie half the time.

So, the answer is 720/2 or 360.

Answer = D


Hi Brent ,

Everything is clear, but can you please explain more above part?

why 720/2?

Thanks,

Sid
Here's an example using 3 people: A, B and C
There are 3! possible outcomes:
BACK -----A B C----FRONT
BACK -----B A C----FRONT
BACK -----B C A----FRONT
BACK -----A C B----FRONT
BACK -----C B A----FRONT
BACK -----C A B----FRONT
Notice that, in the first 3 outcomes, B is behind C
In the last 3 outcomes, C is behind B
So, of the 6 (3!) possible arrangements of A, B and C, HALF have B is behind C, and HALF have C is behind B

The same applies in the original question.
If we just arrange the 6 people in ANY order, we can do so in 720 ways.
In HALF of those 720 arrangements, Frankie will be behind Joey, and in the other HALF, Joey will be behind Frankie

Does that help?

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by hoppycat » Wed May 31, 2017 5:08 am
Matt@VeritasPrep wrote:
gmatbeater1989 wrote:
GMATGuruNY wrote:An alternate approach:

Let the 6 mobsters be A, B, C, D, F and J.
Direction of the line:
Front....Back.

Number of options for A = 6. (Any of the 6 positions.)
Number of options for B = 5. (Any of the 5 remaining positions.)
Number of options for C = 4. (Any of the 4 remaining positions.)
Number of options for D = 3. (Any of the 3 remaining positions.)

Of the two remaining positions, F must occupy the one more to the right, so that he can keep an eye on J.
Number of options for F = 1. (Of the two remaining positions, the one more to the right.)
Number of options for J = 1. (One position left.)

To combine these options, we multiply:
6*5*4*3*1*1 = 360.

The correct answer is D.
nice!
We really want to avoid such an approach if we can, though: the slots method exists to make slot-friendly problems easier, not to make slot-hostile problems more time-consuming!
About the part in red. Is there a way to tell when a question is slot friendly or not?

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by Matt@VeritasPrep » Mon Jun 05, 2017 11:02 pm
hoppycat wrote:About the part in red. Is there a way to tell when a question is slot friendly or not?
That's a great question! Trial and error is typically how I do it, but one sign that slots are a good idea is a set of conditions on where certain people/things can be in the arrangement. (That isn't foolproof, but it often works.)

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by Matt@VeritasPrep » Mon Jun 05, 2017 11:03 pm
GMATsid2016 wrote: Hi Brent ,

Everything is clear, but can you please explain more above part?

why 720/2?

Thanks,

Sid
You could also think about arranging only Frankie and Joey before arranging everyone else. There are only two options (FJ and JF), so half of the arrangements will have FJ and the other half will have JF.