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

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

Love your simplicity Brunel.

I had to write all the combination

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

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

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.

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.

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

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.
- JFAB
- JFBA
- JAFB
- JBFA
- JABF
- JBAF
- AJFB
- BJFA
- AJBF
- BJAF
- ABJF
- BAJF
Done!

Answer = 12 = B

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

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