Three girls and two boys are to be seated in a row. Find the number of ways that this can be done if...

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Smith S.

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Three girls and two boys are to be seated in a row. Find the number of ways that this can be done if...
a) the girls and boys sit alternately
b) a girl sits at each end of the row
c) the girls sit together and the boys sit together.

This is a question on the topic of Permutations and Combinations. I have not understood anything regards to this topic. I would appreciate anyone explaining how Permutations and Combinations work in simpler terms. This, in my opinion, is the most challenging topic in all of Mathematics.
 
Smith S. said:
Three girls and two boys are to be seated in a row. Find the number of ways that this can be done if...
a) the girls and boys sit alternately
b) a girl sits at each end of the row
c) the girls sit together and the boys sit together.

This is a question on the topic of Permutations and Combinations. I have not understood anything regards to this topic. I would appreciate anyone explaining how Permutations and Combinations work in simpler terms. This, in my opinion, is the most challenging topic in all of Mathematics.
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Start with simpler tasks. E.g., there are 5 people named 1,2,3,4,5. How many different ways there are in which they can be ordered/seated ? If this is two difficult try 3 people. Can you do this?
 
Smith S. said:
I believe it's 5! = 120, but I do not know why.
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Your belief is correct. As to why, here are questions/hints:
  1. In how many ways can you pick a person for the first position?
  2. Once you've filled the first position how many ways are there to fill the second position?
  3. What about the third position?
 
Smith S. said:
Three girls and two boys are to be seated in a row. Find the number of ways that this can be done if...
a) the girls and boys sit alternately
b) a girl sits at each end of the row
c) the girls sit together and the boys sit together.

This is a question on the topic of Permutations and Combinations. I have not understood anything regards to this topic. I would appreciate anyone explaining how Permutations and Combinations work in simpler terms. This, in my opinion, is the most challenging topic in all of Mathematics.
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a) [imath](3)(2)(2)(1)(1)=12[/imath] Now I beg you to write out and post the reason why that is correct.
 
a) 3! * 2!. There are three different ways the girls could sit beside a boy, and the boys can sit in two different ways between the girls.
 
Smith S. said:
a) 3! * 2!. There are three different ways the girls could sit beside a boy, and the boys can sit in two different ways between the girls.
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Suppose it had been eight girls and four boys. How many ways can these twelve be arranged in a row so that no two of the boys are next to each other?
 
Smith S. said:
I do not understand.
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O,K. Let start slowly.
Say there are twelve identical ball except four are red & eight are blue.
How many ways can they be arranged in a row if:
1) there are no retractions? answer [imath]\dfrac{12!}{4!\cdot 8!}[/imath] explain!

[imath][/imath][imath][/imath][imath][/imath]

[imath][/imath][imath][/imath]
 
You are removing the same orderings (the denominator). If I switch the position of the red ball with another red ball and a blue ball with another blue ball, I can't really tell the difference since they are identical (meaning the picture would be the same had I taken a snap of it). This limits the arrangement that actually don't matter.
 
Smith S. said:
You are removing the same orderings (the denominator). If I switch the position of the red ball with another red ball and a blue ball with another blue ball, I can't really tell the difference since they are identical (meaning the picture would be the same had I taken a snap of it). This limits the arrangement that actually don't matter.
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The string [imath]RRRRBBBBBBBB[/imath] can arranged in [imath]\dbinom{9}{4}=\dfrac{9!}{4!\times 5!}=126[/imath]
so that none of the [imath]R's[/imath] are together. Can you explain that at all?
HINT: [imath]\underbrace {\quad \;}_1B\underbrace {\quad \;}_2B\underbrace {\quad \;}_3B\underbrace {\quad \;}_4B\underbrace {\quad \;}_5B\underbrace {\quad \;}_6B\underbrace {\quad \;}_7B\underbrace {\quad \;}_8B\underbrace {\quad \;}_9[/imath]
We use the eight [imath]B's[/imath] to create nine places into which to separate the four [imath]R's[/imath].
 
pka said:
The string [imath]RRRRBBBBBBBB[/imath] can arranged in [imath]\dbinom{9}{4}=\dfrac{9!}{4!\times 5!}=126[/imath]
so that none of the [imath]R's[/imath] are together. Can you explain that at all?
HINT: [imath]\underbrace {\quad \;}_1B\underbrace {\quad \;}_2B\underbrace {\quad \;}_3B\underbrace {\quad \;}_4B\underbrace {\quad \;}_5B\underbrace {\quad \;}_6B\underbrace {\quad \;}_7B\underbrace {\quad \;}_8B\underbrace {\quad \;}_9[/imath]
We use the eight [imath]B's[/imath] to create nine places into which to separate the four [imath]R's[/imath].
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I hope you have time to fix this post.
 
Smith S. said:
How did you get 9 and 4 out of 12 letters?
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Thank you for posing a clear question.
Review: How many ways are there to rearrange [imath]RRRRBBBBBBBB[/imath] so that NO TWO of the [imath]R's[/imath] are side by side.
This is a classic occupancy problem. We are using the [imath]B's[/imath] to separate the [imath]R's[/imath] Look again at the following.
[imath]\underbrace {\quad \;}_1B\underbrace {\quad \;}_2B\underbrace {\quad \;}_3B\underbrace {\quad \;}_4B\underbrace {\quad \;}_5B\underbrace {\quad \;}_6B\underbrace {\quad \;}_7B\underbrace {\quad \;}_8B\underbrace {\quad \;}_9[/imath]
The eight [imath]B's[/imath] make nine cells into which we put at most exactly one of the [imath]R's[/imath].
That means we choose four of the nine: [imath]\dbinom{9}{4}=\dfrac{9!}{(4!)(5!)}=126[/imath]. At this point we are done.
For there is no more rearranging to do, because the [imath]B's[/imath] have determined the separating cells
and the [imath]R's[/imath] have been placed into the selected cells.
Example [imath]\underbrace {{\bf R} \;}_1B\underbrace {{\bf R} \;}_2B\underbrace {\quad \;}_3B\underbrace {\quad \;}_4B\underbrace {{\bf R} \;}_5B\underbrace {\quad \;}_6B\underbrace {\quad \;}_7B\underbrace {{\bf R} \;}_8B\underbrace {\quad \;}_9[/imath] in cells 1, 2, 5, & 8.
There are 125 other possible arrangements.

[imath][/imath]
 
How many distinguishable ways can you order a set of 2 distinguishable objects, A and B?

Obviously 2. AB or BA. 2 = 2!

How many distinguishable ways can you order a set of 3 distinguishable objects, A, B, and C?

6. ABC, ACB, BAC, BCA, CAB, CBA. 6 = 3!

How many distinguishable ways can you order a set of 4 distinguishable objects, A, B, C, and D?

Rather than list them all, consider just one of the distinct orderings of three of the four, say BCA. How many places can we slot D in without affecting the order of A, B, and C? There are four slots where we can put D. So 4. DBCA, BDCA, BCDA, and BCAD. But there are 6 distinguishable ways to order A, B, and C. So the answer is 4 * 6 = 4 * 3! = 4! = 24.

You should now understand WHY the number of distinguishable ways you can order a set of n distinguishable objects is n!. So in word problems look for the concept of “distinct.” That word may not be present, but if the concept is there, understand its importance. Similarly look for the concept of order, whether it is explicitly present or not.

Let’s assume that there are female triplets all dressed alike and male twins dressed alike. How many distinguishable ways can they be seated in a line so that the three females are seated adjacently. Think about the seats being numbered. We could put the sisters in seats 123, 234, or 345. The answer is 3. Notice that if we seat the sisters in seats 234, then the twin boys are not seated adjacently. So if the problem is to seat the twin brothers adjacently and the triplet sisters adjacently, there are only 2 distinguishable seatings, BBGGG or GGGBB. Now relax the assumption that the girls all look alike.. How many different ways can we distinguish the girls? 3! Now we have 2 * 3! = 12. Now relax the assumption that the boys look alike. How many ways can we distinguish the boys? 2!. So our answer is 2 * 3! * 2! That happens to be 24, which happens to be 4!. But you can can find the answer through logic to be 2 * 3! * 2!. The answer of 4! does not make the logic clear.

The formulas [imath]n![/imath] and [imath]\dfrac{n!}{k! * (n - k)!}[/imath] are easy to remember. But counting problems requires very careful thinking, not reliance on a few formulas. It frequently helps to break the problem down into easier problems.
 
Steven G said:
That is why the 9 should be an 8.
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@steven, did you study the diagram? [imath]\underbrace {\quad \;}_1B\underbrace {\quad \;}_2B\underbrace {\quad \;}_3B\underbrace {\quad \;}_4B\underbrace {\quad \;}_5B\underbrace {\quad \;}_6B\underbrace {\quad \;}_7B\underbrace {\quad \;}_8B\underbrace {\quad \;}_9[/imath]
In counting questions, in particular separator problems, [imath]N[/imath] separators create [imath]N+1[/imath] cells.
Because we have eight B's as separators & [imath]8+1=9[/imath] that is why the 9 is correct.


[imath][/imath]
 
pka said:
@steven, did you study the diagram? [imath]\underbrace {\quad \;}_1B\underbrace {\quad \;}_2B\underbrace {\quad \;}_3B\underbrace {\quad \;}_4B\underbrace {\quad \;}_5B\underbrace {\quad \;}_6B\underbrace {\quad \;}_7B\underbrace {\quad \;}_8B\underbrace {\quad \;}_9[/imath]
In counting questions, in particular separator problems, [imath]N[/imath] separators create [imath]N+1[/imath] cells.
Because we have eight B's as separators & [imath]8+1=9[/imath] that is why the 9 is correct.


[imath][/imath]
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pka said:
The string [imath]RRRRBBBBBBBB[/imath] can arranged in [imath]\dbinom{9}{4}=\dfrac{9!}{4!\times 5!}=126[/imath]
so that none of the [imath]R's[/imath] are together. Can you explain that at all?
Click to expand...
I am sorry. I did not realize that what is in red was part of the prior line.
 
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