SOLUTION: April, Bill , Candace, and Bobby are to be seated at random in a row of 8 chairs. What is the probability that April and Bobby will occupy the seats at the end of the row?

Algebra ->  Permutations -> SOLUTION: April, Bill , Candace, and Bobby are to be seated at random in a row of 8 chairs. What is the probability that April and Bobby will occupy the seats at the end of the row?      Log On


   

Question 1207172: April, Bill , Candace, and Bobby are to be seated at random in a row of
8 chairs. What is the probability that April and Bobby will occupy the seats at the end of the row?
Found 6 solutions by chimichurri, AnlytcPhil, ikleyn, Edwin McCravy, greenestamps, Plocharczyk:
Answer by chimichurri(1) About Me  (Show Source):
Answer by AnlytcPhil(1806) About Me  (Show Source):
You can put this solution on YOUR website!

Here is an alternate way to solve the problem using stars and bars.  

Let's suppose the correct interpretation for a successful arrangement is for one
of the pair A and B, to be in the far left chair, and the other to be in the far
right chair.

There are 2 ways to arrange A and B on the ends.

There are 6*5=30 ways to arrange C and D between them in the 6 chairs between
them.

So the number of SUCCESSFUL arrangements is 2*6*5=60.

Now let's look at the number of POSSIBLE arrangements. Take this sample

CDBA.  There are 4!=24 arrangements like this one.

We must now insert 4 empty chairs among them. 

This involves the number of partitions of 4 of length 5.

We make a row of 4 stars and 4 bars, like this sample:

**||*|*|

That is the partition 2+0+1+1+0=4.  That's because 
there are: 
2 stars left of the 1st bar, 
0 stars between the 1st and 2nd bar,
1 star between the 2nd and 3rd bar,
1 star between the 3rd and 4th bar, and
0 stars after the 4th bar.

This sample partition of 4 of length 5, which is 2+0+1+1+0 makes the sample 
ADCB to become:
 
_ _ A D _ C _ B

where the blanks represent the 4 empty chairs.

Now we find the number of ways to insert the 4 empty chairs among the 4 people.

There are 4 indistinguishable stars and 4 indistinguishable bars.

So there are 8%21%2F%284%214%21%29%22%22=%22%2270 ways to put 4 empty chairs 

among the 4 people.

So let's put what we've said here all together: 

There are 4!=24 ways to form something like this: ADCB.

Then there are 70 ways to form something like this _ _ A D _ C _ B

That's 24x70 = 1680 possible ways.

So the probability is 60%2F1680%22%22=%22%221%2F28

Edwin

Answer by ikleyn(52780) About Me  (Show Source):
You can put this solution on YOUR website!
.
April, Bill , Candace, and Bobby are to be seated at random in a row of
8 chairs. What is the probability that April and Bobby will occupy
the seats at the end of the row?
~~~~~~~~~~~~~~~~~ 

April   can occupy any of 8 seats;
Bill    can occupy any of 7 remaining seats;
Candace can occupy any of 6 remaining seats;
Bobby   can occupy any of 5 remaining seats.


In all, there are 8*7*6*5 = 1680 possible different placements.


The favorable placements are 

    (a)  April is at the most left seat; Bobby is at the most right seat;
         two other persons in any 2 of 6 remaining seats,
         which gives 6*5 = 30 options.


    (b)  April is at the most right seat; Bobby is at the most left seat;
         two other persons in any 2 of 6 remaining seats,
         which gives 6*5 = 30 other options.


Therefore, the desired probability is  P = %2830%2B30%29%2F1680 = 60%2F1680 = 1%2F28.


ANSWER.  The desired probability is 1/28.

Solved.

----------------------

The problem's formulation is not totally clear and admits different interpretations.

My interpretation was that April and Bobby occupy the seats at two opposite ends of the row.

Other possible interpretation can be that April and Bobby occupy two adjacent seats at some of the two ends.

When a problem admits different interpretations, it is a fault of its creator.



Answer by Edwin McCravy(20054) About Me  (Show Source):
You can put this solution on YOUR website!

Here is an alternate way to solve the problem using stars and bars.  

Let's suppose the correct interpretation for a successful arrangement is for one
of the pair A and B, to be in the far left chair, and the other to be in the far
right chair.

There are 2 ways to arrange A and B on the ends.

There are 6*5=30 ways to arrange C and D between them in the 6 chairs between
them.

So the number of SUCCESSFUL arrangements is 2*6*5=60.

Now let's look at the number of POSSIBLE arrangements. Take this sample

CDBA.  There are 4!=24 arrangements like this one.

We must now insert 4 empty chairs among them. 

This involves the number of partitions of 4 of length 5.

We make a row of 4 stars and 4 bars, like this sample:

**||*|*|

That is the partition 2+0+1+1+0=4.  That's because 
there are: 
2 stars left of the 1st bar, 
0 stars between the 1st and 2nd bar,
1 star between the 2nd and 3rd bar,
1 star between the 3rd and 4th bar, and
0 stars after the 4th bar.

This sample partition of 4 of length 5, which is 2+0+1+1+0 makes the sample 
ADCB to become:
 
_ _ A D _ C _ B

where the blanks represent the 4 empty chairs.

Now we find the number of ways to insert the 4 empty chairs among the 4 people.

There are 4 indistinguishable stars and 4 indistinguishable bars.

So there are 8%21%2F%284%214%21%29%22%22=%22%2270 ways to put 4 empty chairs 

among the 4 people.

So let's put what we've said here all together: 

There are 4!=24 ways to form something like this: ADCB.

Then there are 70 ways to form something like this _ _ A D _ C _ B

That's 24x70 = 1680 possible ways.

So the probability is 60%2F1680%22%22=%22%221%2F28

Edwin

Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


The immediate interpretation, upon reading the problem, is that April and Bobby occupy the two seats at opposite ends of the row.

But admittedly there are other possible interpretations.

In this response, I assume that April and Bobby occupy the two seats at opposite ends of the row.

In that case, the positions of the other people is irrelevant. It doesn't even matter how many other people there are (as long as the total number of people is not more than 8). The only thing that matters is that April and Bobby occupy the two seats on opposite ends of the row.

There are two possibilities:
(1) April in seat 1 and Bobby in seat 8. The probability that April sits in seat 1 is 1/8; after that, the probability that Bobby sits in seat 8 is 1/7. So the probability is (1/8)*(1/7) = 1/56.
(2) Bobby in seat 1 and April in seat 8. The probability that Bobby sits in seat 1 is 1/8; after that, the probability that April sits in seat 8 is 1/7. So the probability is (1/8)*(1/7) = 1/56.

So the probability that April and Bobby occupy the two seats on opposite ends of the row is 1/56+1/56 = 2/56 = 1/28.

Answer by Plocharczyk(17) About Me  (Show Source):
You can put this solution on YOUR website!

Here is an alternate way to solve the problem using stars and bars.  

Let's suppose the correct interpretation for a successful arrangement is for one
of the pair A and B, to be in the far left chair, and the other to be in the far
right chair.

There are 2 ways to arrange A and B on the ends.

There are 6*5=30 ways to arrange C and D between them in the 6 chairs between
them.

So the number of SUCCESSFUL arrangements is 2*6*5=60.

Now let's look at the number of POSSIBLE arrangements. Take this sample

CDBA.  There are 4!=24 arrangements like this one.

We must now insert 4 empty chairs among them. 

This involves the number of partitions of 4 of length 5.

We make a row of 4 stars and 4 bars, like this sample:

**||*|*|

That is the partition 2+0+1+1+0=4.  That's because 
there are: 
2 stars left of the 1st bar, 
0 stars between the 1st and 2nd bar,
1 star between the 2nd and 3rd bar,
1 star between the 3rd and 4th bar, and
0 stars after the 4th bar.

This sample partition of 4 of length 5, which is 2+0+1+1+0 makes the sample 
ADCB to become:
 
_ _ A D _ C _ B

where the blanks represent the 4 empty chairs.

Now we find the number of ways to insert the 4 empty chairs among the 4 people.

There are 4 indistinguishable stars and 4 indistinguishable bars.

So there are 8%21%2F%284%214%21%29%22%22=%22%2270 ways to put 4 empty chairs 

among the 4 people.

So let's put what we've said here all together: 

There are 4!=24 ways to form something like this: ADCB.

Then there are 70 ways to form something like this _ _ A D _ C _ B

That's 24x70 = 1680 possible ways.

So the probability is 60%2F1680%22%22=%22%221%2F28

Edwin