A family of 2 parents and 2 children is waiting in line to order at a

Question

A family of 2 parents and 2 children is waiting in line to order at a fast-food restaurant. Joey, the younger child, has a tendency to cause mischief when he is not watched carefully. Because of this, the father wants to keep Joey ahead of him in line at all times. How many different ways can the family arrange themselves in line such that the father is able to watch Joey?

A. 6
B. 12
C. 24
D. 36
E. 72

A. 6
B. 12
C. 24
D. 36
E. 72

Bunuel · Accepted Answer

The total number of ways to arrange 4 people is 4! = 24. In half of these arrangements, the father will be behind Joey, and in the other half, Joey will be behind his father. Therefore, the answer is 4!/2 = 12. Answer: B. Questions about the same concept to practice: https://gmatclub.com/forum/susan-john-d ... 30743.html https://gmatclub.com/forum/meg-and-bob- ... 58095.html https://gmatclub.com/forum/six-mobsters ... 26151.html https://gmatclub.com/forum/mary-and-joe ... 26407.html https://gmatclub.com/forum/goldenrod-an ... 82214.html https://gmatclub.com/forum/in-how-many- ... 91460.html Hope it helps.

shreeny · Answer

Thank You.
I interpreted the question as "father wants to keep Joey  right  ahead of him in line" and got the answer as 6.

Bunuel · Answer

Yes, in this case the answer would be 6: {M}{C}{JF} --> 3!=6.

surya167 · Answer

Love your simplicity Brunel.

I had to write all the combination

MFCJ
MFJC
FMCJ... etc.. counting all would give us 12

theGame001 · Answer

Bunuel, can you explain how do you conclude that half of the cases he will be ahead? Is it something to do with even numbers of people?

Thanks

Bunuel · Answer

I guess you didn't follow the links my previous post.

Consider this: no matter how this 4 will be arranged there can be only two scenarios, either father 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 father. How else? Now, ask yourself: why should one arrangement give more ways than the other?

Hope it's clear.

Nevernevergiveup · Answer

Positions  1st   2nd   3rd   4rd

The cases are
 1. If Joe is in first position then his father can occupy any of the remaining 3 positions.
2. If Joe is in second position then his father can occupy any of the 3rd and 4th position in the order.
3. If Joe is in third position then his father can occupy only last position.
 
In each case the other child and parent can occupy any of the remaining 2 positions in  2!  ways.

different ways as per the requirement =  (3+2+1)2!=6*2=12

option B is the answer.

JeffTargetTestPrep · Answer

We are given that 2 parents and 2 children are waiting in line, and we must determine in how many ways they can line up with Joey ahead of the father at all times. To determine the number of ways we can refer to the following equation:

Total number of ways to arrange the group = # ways with Joey ahead of father + # ways with Joey behind father.

Since the total number of ways in which Joey could be ahead of his father is equal to the total number of ways in which Joey could be behind his father, and since the total number of ways to arrange the group is 4P4 = 4! = 24, the number of ways in which Joey could be ahead of his father is 24/2 = 12 ways.

Answer: B

BrentGMATPrepNow · Answer

Bunuel's and Jeff's approaches are great. I just wanted to point out that, when the answer choices are RELATIVELY SMALL (as they are here), you should also consider LISTING AND COUNTING as one of your approaches, especially if you don't identify any other approaches. In many cases, the simple process of listing outcomes will help us gain valuable insight into a fast way to solve the question.

So, if we let   J   = Joey,   F   = Father, and let A and B equal the two other people, we can start listing possible arrangements in a  systematic  way.

FRONT......BACK.
-   J    F  AB
-   J    F  BA
-   J  A  F  B
-   J  B  F  A
-   J  AB  F  
-   J  BA  F  
- A  J    F  B
- B  J    F  A
- A  J  B  F  
- B  J  A  F  
- AB  J    F  
- BA  J    F  
Done!

Answer = 12 = B

Even though this approach is the least mathematical, we can easily solve the question in under two minutes.

RELATED VIDEO
 https://www.youtube.com/watch?v=KGvBUx3FUWQ

rohansirohia · Answer

I will be grateful if someone can clarify why my approach is giving an incorrect answer :

If I need to keep Joey( J ) ahead of Father ( F ), I will have the following arrangement :

_ _ J _ _ F _ _

Each `_` is a space where I can place the remaining 2 people and then permute them, so total will be 6C2 * (2!).

Why is this wrong?

Bunuel · Answer

1 2 J 3 4 F 5 6

This approach will result in duplicates. For example, if positions 1 and 3 are chosen, you get X - J - Y - F, and if positions 1 and 4 are chosen, you still get the same arrangement: X - J - Y - F. Please review the solutions provided above. This question has a much simpler solution.

sujoykrdatta · Answer

Let us name them: F (father), J (Joey), A (other adult), C (other child)
We need to arrange them so that J is ahead of F, but not necessarily together or immediately ahead

Method 1: Total arrangements = 4! = 24
However, we need to remove the arrangements of F and J (which can be done in 2! = 2 ways and is included in 24 as a factor)
So, the answer = 4!/2! = 24/2 = 12

Method 2: 
There are 4 positions. Choose any 2 for J and F, but do not arrange them since J must remain ahead of F. 
This can be done in 4C2 = 6 ways
Arrange A and C in 2! = 2 ways
Thus, total ways = 4C2 * 2! = 6 * 2 = 12

Answer B

jacob46 · Answer

Hi Bunuel,

When I was solving the question, I did the following; 
Since there are 4 in total but the  son is always ahead of father . I took them as 3 slots Father and Son together as one and 2 for the rest 2 arriving at 3!. I did not again further multiply it by 2! because there is only one arrangement for the father and son hence it would be 3! x 1, arriving at the answer 6. Could you please help me understand where my logic is flawed?

Thank you

Bunuel · Answer

"Ahead of him in line" does not necessarily mean directly ahead, Joey just needs to be somewhere earlier in the line than the father. They can have other people standing between them. Thus, you cannot group them as one block.

Also, please check the links given in my earlier post to practice similar questions for better understanding.

luisdicampo · Answer

Deconstructing the Question  
There are 4 family members.
Total possible arrangements:
 4!=24

Constraint: Joey must stand ahead of the father.

Step-by-step  
For any arrangement where the father is ahead of Joey,
there is a corresponding arrangement where their positions are swapped.

Thus exactly half of all arrangements satisfy the condition.

\frac{24}{2}=12

Answer: B

A family of 2 parents and 2 children is waiting in line to order at a

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