Problem Solving

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

I strategically guessed the correct answer however I have one doubt. My approach was to put Joey in certain place in a line and count desirable outcomes. So, when the Jeoy is last in the raw no outcomes.

J _ _ _ no outcomes
_ J _ _ father must be behind him (to the left of the letter J) and therefore leaving 2 x 1 for the rest of the line = 2
_ _ J _ this is the part where I stuck. How to calculate desirable outcomes in this position?
_ _ _ J all possible ways meaning 3 x 2 x 1 = 6

OA is B

Milovan wrote: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

Let the line look as follows:
FRONT......BACK.
Let J=joey and F=father.
In a viable arrangement, J is to the left of F, allowing the father to keep an eye on Joey.

The number of ways to arrange the 4 people = 4! = 24.
In any given arrangement, the probability that J is to the left of F is the same as the probability that F is to the left of J.
Thus:
In 1/2 of the arrangements, J is to the left of F.
In the other 1/2 of the arrangements, F is to the left of J.
Thus:
The number of arrangements in which J is to the left of F = (1/2)(24) =12.

The correct answer is B.

An alternate approach is to count all of the arrangements in which J is to the left of F.

Case 1: J _ _ _
Number of ways to arrange the remaining 3 people = 3! = 6.

Case 2: _ J _ _
Number of options for the leftmost position = 2. (Anyone but F.)
Number of ways to arrange the remaining 2 people = 2! = 2.
To combine these options, we multiply:
2*2 = 4.

Case 3: _ _ J _
Number of options for the rightmost position = 1. (Must be F.)
Number of ways to arrange the remaining 2 people = 2! = 2.
To combine these options, we multiply:
1*2 = 2.

Total ways = 6+4+2 = 12.

A third approach is to WRITE OUT all of the possible arrangements.
Let J=Joey, F=father, M=mother, X=other child.

Case 1: J _ _ _
JFMX
JFXM
JMFX
JMXF
JXFM
JXMF
6 options.

Case 2:_ J _ _
MJFX
MJXF
XJFM
XJMF
4 options.

Case 3: _ _ J _
MXJF
XMJF
2 options.

Total ways = 6+4+2 = 12.

Milovan wrote: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

I strategically guessed the correct answer however I have one doubt. My approach was to put Joey in certain place in a line and count desirable outcomes. So, when the Jeoy is last in the raw no outcomes.

J _ _ _ no outcomes
_ J _ _ father must be behind him (to the left of the letter J) and therefore leaving 2 x 1 for the rest of the line = 2
_ _ J _ this is the part where I stuck. How to calculate desirable outcomes in this position?
_ _ _ J all possible ways meaning 3 x 2 x 1 = 6

OA is B

There is one more approach that i learnt from Mitch from one of the posts on P&C that he has posted on this forum. The approach is following:

Let the four people be: J(Joey), F(Father), A, B.
There are 4 positions to be filled. Out of which A and B can occupy any of the four positions.

Therefore, # of options for A = 4
and # of options for B = 3

Now, since J must stand before F,
Thus, # of options for J = 1(There is only one way that J may occupy the position before F)
and # of options for F = 1

Combining all the options,
3*4*1*1 = 12.

Thanks to all of you.

The stuck factor was the fundamental of permutations and that is to start with the most restrictive place in the Case 2! I should started with the left position (in front of the kid) and not two right positions (behind the kid). Thanks to Mitch I spotted my mistake.

By the way Parveen10 nice trick you showed taken from Mitch.