Six mobsters have arrived at the theater for the premiere of

Questions about the same concept to practice:
https://gmatclub.com/forum/susan-john-da ... 30743.html
https://gmatclub.com/forum/meg-and-bob-a ... 58095.html
https://gmatclub.com/forum/in-how-many-d ... 91460.html
https://gmatclub.com/forum/mary-and-joe- ... 26407.html
https://gmatclub.com/forum/goldenrod-and ... 82214.html

Hope it helps.

This is exactly what has been taken into account. Consider this: no matter how this 6 will be arranged there can be only two scenarios, either Frankie is behind Joey (when saying behind I mean not just right behind but simply behind, there may be any number of persons between them) OR Joey is behind Frankie. Well, this is pretty obvious. So, as 6 can be arranged in 6! ways, so half of this 6! ways will have *F*J* and half *J*F*. 6!/2=360.

Hope it's clear.

Bunuel,

Thanks for your reply. A lot of GMAT questions are tricky in their wording. This can cause me to OVERanalyze a passage.

Heres how i worked it out:

6! = Total possible combinations
So half of, Joey is in front and the other half Frankie is.

6!/2 = 320 = The half where Joey is in front and Frankie is in back.

Then it states:
But that doesn't count "insists upon standing behind Joey in line at the concession stand, though not necessarily right behind him"

Heres my interpretation of that statement:

It can't be x,x,x,x,F,J
Rather it can only be x,x,x,F,x,J

I don't know... I maybe goin crazy !!!

"not necessarily right behind him" means that he may or may not be right behind him. It's not necessary Frankie to be right behind Joey but if this happens still no problem.

So, **F**J good as well as **FJ**.

man you are a genius !!

you look at the problem from a very simple angle..

Here's the answer.

There are fifteen ways F would be behind J all the time. Think of it this way. There are six places, and you have to select two places for F & J, and where the other four's positions does not matter. This can be put into numbers as 6!/4!2! giving us 15.

So you'd have the following:

FJ????, F?J???, F??J??, F???J?, F????J
?FJ???, ?F?J??, ?F??J?, ?F???J
??FJ??, ??F?J?, ??F??J
???FJ?, ???F?J
????FJ

15 scenarios in all.

Now, you can use counting techniques to get all the possible orders of the four remaining mobsters. That would be 4! = 24.

Multiply the 15 cases with all of the possible orders of the four remaining mobsters - 15*24 = 360.

Hi,

maybe this will help:

take 2 objects, x and y. How many ways to arrange them?

Clearly, 2: xy and yx. Notice that in half the arrangements (1 out of 2) x is before y, and in the other half y is before x.

Take 3 objects, x, w, and y. There are 3! or 6 arrangements. In half (ie, 3) of those arrangements, x is before w, while in the other half w before x. Likewise, in half of the arrangements x is before y while in the other half y before x. And, also, in half the arrangements, w is before y while in the other half y before w.

Why would it be the case that Joey can be arranged ahead of Frankie more or less often than Frankie can be arranged ahead of Joey? Why not the other way around?

So, here, the easiest way to solve is certainly to take just half of the total arrangements: 6!/2 = 360, and choose D.



Yes, it certainly can as the above poster demonstrated!

But many combinatorics questions on the GMAT resist pure formulaic treatment. A little bit of reasoning on these questions can save you an immense amount of time!

The important thing to realize here is that Frankie and Joey are absolutely identical elements of this arrangement.
Say, I have 3 elements: A, B and C
I can arrange them in 3! ways:

ABC
ACB
BAC
BCA
CAB
CBA
Look at them carefully. In 3 of them A is before B and in other 3, B is before A. It will be this way because A and B are equal elements. There is no reason why A should be before B in more cases than B before A. Similarly, you can compare B and C or A and C.
Hence, when we arrange all 6 people in 6! ways, in half of them Frankie will be before Joey and in other half, Joey will be before Frankie.

Another way that I tried to solve..may be lengthier too...

5! + 4(4!) + 3 (4!) + 2 (4!) + 1(4!) = 360

Assume that in the first case: that XXXXXJ - Frankie total number of people can be arranged in 5! ways.. <Frankie will be always behind Joey>

Second case: XXXXJX - Given that F has to behind J..if F is fixed at 4th then remaining can be arranged in 4! ways and this has to repeated 4 times... so 4(4!)..same has to be done when F is in third position and so on...

I hope it helps!

Here's how i did it. Hope it helps someone.

Since F has to be behind J.

It would be better if you write them ("JXXXXF") vertically, i am doing horizontally to save space.

JXXXXF JXXXFX JXXFXX JXFXXX JFXXXX

24 Ways 24 Ways 24 Ways 24 Ways 24 Ways = 120 Ways

(24 ways because 4X's could be rearranged in 4! ways)

XJXXXF XJXXFX XJXFXX XJFXXX

24 Ways 24 Ways 24 Ways 24 Ways = 96 Ways

XXJXXF XXJXFX XXJFXX

24 Ways 24 Ways 24 Ways = 72 Ways

XXXJXF XXXJFX

24 Ways 24 Ways = 48 Ways

XXXXJF

24 Ways = 24 Ways


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

Now the method I'm discussing has already been shared, but the explanations were not too clear for me, so sharing a detailed explanation.
Now following can be the available arrangements (please note that the question requires us to place Frankie behind Joey in line, though not necessarily right behind him):
_ _ _ _ _ F: when F is at the last, there are 5 options for J and the rest can be arranged in 4! ways: 5x4!
_ _ _ _ F _: when F is at the second last position, there are 4 options for J and the rest can be arranged in 4! ways: 4x4!
_ _ _ F _ _: when F is at the third last position, there are 3 options for J and the rest can be arranged in 4! ways: 3x4!
_ _ F _ _ _: when F is at the third third spot from the front, there are 2 options for J and the rest can be arranged in 4! ways: 2x4!
_ F _ _ _: when F is at the third third spot from the front, there is only 1 options for J and the rest can be arranged in 4! ways: 4!
Now F can not be placed at the first spot as he has to be behind J, so above are al the situations possible and the sum would give us the total arrangements.
total arrangements = 5x4! + 4x4! + 3x4! + 2x4! + 4! = (5+4+3+2+1)4! = 15x4! = 360

Bunuel

My take -

Total 6 places to fill.

_ _ _ _ _ J. ( No. Of place for F is 5 OR F can take any place among the 5)

_ _ _ _ J _ .( Places for F = 4)
Similarly
F will have 3,2,1 places as per J position.
Total ways = 5*4*3*2*1=120.

Where did I go wrong? Please explain !
chetan2u

Posted from my mobile device

You are not taking the combinations in which others can stand...
Say I wanted to do as per your way..
1) _,_,_,_,_, J
So remaining 5 in 5!
2) _,_,_,_, J,_
F can take any of 4, and the remaining 4 seats can be taken in 4! Ways
So 4*4!
3) _,_,_, J,_,_
Similarly 3*4!
4) _,_, J,_,_,_
So 2*4!
5) _, J,_,_,_,_
So 1*4!

Total
5!+4*4!+3*4!+2*4!+1*4!=4!(5+4+3+2+1)=24*15=360..

Hope it helps

We see that we can arrange the 6 mobsters in 6! = 720 ways. Since the number of ways in which Joey stands in front of Frankie is equal to the number of ways he stands behind him, the number of ways that the group can be arranged to satisfy Frankie’s requirement is 720/2 = 360 ways.

Answer: D

Can you please tell me where I am going wrong here? Since F & J must be together, we only have 5! ways to arrange this line. And we need to take into account that F can be ahead or behind J. So, I approached this question by saying that (5!)*(2!) = 240. None of the answer choices fit my answer. Where did I go wrong?

VeritasKarishma @Bunnel ScottTargetTestPrep JeffTargetTestPrep

shenwenlim

Frankie and Joey needn't be together. We are given that Frankie wants to stand behind Joey but NOT necessarily right behind him. So Joey could stand at number 1 position in the line and Frankie could take any of 2 or 3 or 4 or 5 or 6 positions. If Joey stands at number 2 position, Frankie could take number 3 or 4 or 5 or 6. etc.

It is a problem involving symmetry.

6 people can stand in a line in 6! ways.
In half of these 6! ways, Joey will be ahead of Frankie. In the other half Frankie will be ahead of Joey (for us, Joey and Frankie are equivalent elements).

So in 6!/2 = 360 ways, Joey will be ahead of Frankie. These are our acceptable cases.

For more, check this: https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2011/1 ... s-part-ii/

Hi Bunuel, how were you able to provide this list of questions on the same concept? If I were to practice some questions of Quant/Verbal of any given concept, how and where may I search so?

I am new to this website. Any one who is willing to help, your answers are welcome.

Thanks in advance­