SOLUTION: I have five children and want to place them in a line for a photograph. However, Hugh refuses to stand anywhere in between Louise and Richard. How many ways are there to place th

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Question 1154710: I have five children and want to place them in a line for a photograph.
However, Hugh refuses to stand anywhere in between Louise
and Richard. How many ways are there to place the children in a line
and still keep Hugh happy?
Found 2 solutions by greenestamps, Edwin McCravy:
Answer by greenestamps(13200)   (Show Source): You can put this solution on YOUR website!


Designate the 5 children with H (Hugo), L (Louise), R (Richard), and A and B (unnamed). We need to find the number of arrangements of those 5 letters in which H is not between L and R.

Here is one relatively fast and efficient way to find the answer.

Consider separate cases for each place in line that Hugo can stand.

(1) Hugo first: Hxxxx
The other 4 can be in any order: # ways = 4! 24

(2) Hugo second: xHxxx
The person to the left of Hugo must be A or B (2 choices); the three to his right can be in any order. # ways = 2*3! = 12

(3) Hugo in the middle: xxHxx
Louise and Richard need to be either both to the left or both to the right (2 choices). The pair on the left of Hugo can be in either of 2 orders; likewise for the pair on the right. # ways = 2*2*2 = 8

(4) Hugo 4th: xxxHx
By symmetry, this is the same as if he were second. # ways = 12

(5) Hugo last: xxxxH
Again by symmetry, this is the same as if he were first. #ways = 24

Total number of ways the children can stand in a line without Hugo being between Louise and Richard: 24+12+8+12+24 = 80.

This problem is a good one for practicing logical analysis. To get further practice, you could do a similar analysis to find the number of ways of arranging the children so the Hugo IS between Louise and Richard. Since the total number of ways of arranging the 5 children is 5!=120, that number should be 40.

And there are possibly easier ways to find the answer that are completely different than this....

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NOTE: See the response from tutor @Edwin for an analysis that shows there are indeed 40 arrangements where Hugo IS between Louise and Richard.

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I found myself thinking about this problem as I was waking up this morning, and I came up with a couple of faster ways to get to the answer. Here they are....

Both these ways find the answer by finding the number of ways the children can be lined up in which Hugo IS between Louise and Richard.

The first way uses two cases which are the same by symmetry.

Place the three children Louise, Hugo, and Richard in a line in that order: LHR.

Now place either of the other two children in the line. There are three children currently in the line, so there are four choices for where that child can go -- 1st, 2nd, 3rd, or 4th.

Now similarly place the last child in the line. Now, since there are currently four children in the line, there are five choices for where this last child can go.

Then the total number of ways the five children can stand in the line with Louise, Hugo, and Richard in that order is 4*5=20.

And the analysis is exactly the same if the three named children are in the order RHL.

So there are 40 ways the children can stand in the line with Hugo between Louise and Richard. Then, since there are 120 ways total that the five children can stand in the line, there are 80 ways for them to line up in which Hugo is NOT between Louise and Richard.

And here finally is a path that gets you to the answer even faster, using probability, instead of counting arrangements.

Start by placing Hugo in the line.
Next take either Louise or Richard and have that child stand next to Hugo, on either side.
Last place the other of Louise and Richard in the line. Since there are currently two children in line, there are three choices of where this third child can stand. And only one of those three choices will put Hugo between Louise and Richard.

So there is a probability of 1/3 that Hugo ends up between Louise and Richard.

And so, of the 120 possible arrangements of the 5 children in a line, 40 of them have Hugo between Louise and Richard; so 80 of the arrangements will make Hugo happy.
Answer by Edwin McCravy(20055)   (Show Source): You can put this solution on YOUR website!
Let the 5 children be Hugh, Louise, Richard, Ann, and Bob.
Let their initials stand for them.


There would be 5! = 5∙4∙3∙2∙1 = 120 ways if there were no restrictions.

Let's enumerate the unwanted cases where H is between L and R, and subtract
them from the 120.

There are 2 ways to arrange them with Hugh between them. We can have _L_H_R_
or _R_H_L_, where the blanks can contain 0, 1 or 2 people. 

For each of those 2 ways, there are two cases:

Case 1: A and B are not side by side.

There are 4 blanks to put A in, and 3 remaining blanks to put B in.
That's 4∙3=12 for Case 1.

Case 2: A and B are side-by-side
They can be side-by-side 2 ways, AB or BA
For either of those 2 ways, there are 4 places to put them.
That's 2∙4=8 ways for Case 2.

The total for those two cases is 12+8=20.

So the number of unwanted arrangements is 2∙(12+8)=2∙20 = 40 unwanted cases.

We subtract that from the 120 and so the answer is 120-40 = 80.

Edwin
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