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

[khansaheb]

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.

Let us try to do part c) first.

Group the 3 girls as G and group 2 boys as B.

How many ways can you arrange these groups in a line → GB and BG → in 2! ways

In group G, we have 3 distinguishable girls (assumption) G₁,G₂ and G₃. Those can be arranged as G₁G₂G₃ or G₁G₃G₂ or ...... 3! ways

In group B, we have 2 distinguishable boys (assumption) B₁,B₂ . Those can be arranged as B₁,B₂ or B₂,B₁ or ...... 2! ways

So - how many ways can you arrange 3 girls and 2 boys in a line such that the girls sit together and the boys sit together.

Please share your calculations....

 

[pka]

If the girls are [imath]A, C, D[/imath] and the boys are [imath]X,Y[/imath].
The set [imath]G=\{ACD,ADC,CAD,CDA,DAC,DCA\}[/imath] is the set of all the ways the girls sit together.
The set [imath]B=\{XY,YX\}[/imath] is the set of all the ways the boys sit together.

The cross product [imath]G\times B[/imath] contains twelve ordered pairs, ex:[imath](CAD,YX)\in G\times B[/imath]
Each pair represents a way that the group of girls are sitting together followed the group of boys sitting together.

The cross product [imath]B\times G[/imath] contains twelve ordered pairs, ex:[imath](XY, ADC)\in B\times G[/imath]
Each pair represents a way the group of boys are sitting together followed the group of girls sitting together.

Thus what is the answer to (c)?

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

 

[Smith S.]

a) 12
b) 36
c) 24

I used the concept of Permutation here.

 

[Smith S.]

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?

8! * 9P4. Is this correct?

 

[Smith S.]

Is this a combination or permutation question: "A football club has 30 players. In how many different ways can a captain and a vice-captain be
selected at random from these players?"? Should not it be a combination question?

 

[pka]

8! * 9P4. Is this correct?

One last time. We have eight girls and four boys.
We want the seat them in a a row so that no two boys are adjacent.
The girls are separators, the 8 create 8+1=9 places to seat a boy.
First select the four places: [imath]\dbinom{9}{4}=126[/imath].
Second: the boys can be arranged in [imath]4!=24[/imath] ways.
Third: the girls can be arranged in [imath]8!=40320[/imath] ways.
Now multiply.

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

 

[NotJeffM]

Is this a combination or permutation question: "A football club has 30 players. In how many different ways can a captain and a vice-captain be
selected at random from these players?"? Should not it be a combination question?

You seem to be under the impression that all there is to this field is deciding whether to use the combination formula or the permutation formula. You may have to use different formulas and put them together in a logical way. It is not a matter of saying it is formula C or else formula P.

 

[Smith S.]

You may have to use different formulas and put them together in a logical way.

How can I possibly do that if I do not understand the logic behind it? I was hoping you could guide me to some sources I can read and understand from.

 

[NotJeffM]

How can I possibly do that if I do not understand the logic behind it? I was hoping you could guide me to some sources I can read and understand from.

People like PKA know more formulas that deal with specific types of problems, but it is highly unlikely that in the course you are studying you will come across enough such problems to make it remotely sensible to try memorizing a bunch of additional formulas.

Did you read post 15? Did you read post 21? Did you read post 26? There you were shown how people broke your problem down into a group of simpler problems that use the basic formulas you have been taught.

What is usually taught as permutations and combinations is two or sometimes three formulas that need to be put together in a logical way to solve a problem.

How many ways can you pick one person to be captain out of thirty team members? See what I have done? I broke the problem down into parts. Now the answer to that question is self-evidently THIRTY. Technically, you are correct that that must be a combination problem because we cannot put into distinct orders a single item.

[math]\dfrac{30!}{1! * (30 - 1)!} = \dfrac {30 * 29!}{1 * 29!} = \dfrac{30}{1} = 30.[/math]
OK, but that does not answer the entire problem. We now have to solve some remaining smaller problems to solve the entire problem. Once we have chosen a captain, what is the size of the remaining pool from which to choose a vice-captain? That is neither a permutation nor a combination problem; it is a simple arithmetic problem, namely

[math]30 - 1 = 29.[/math]
So how many ways can you choose a vice-captain from the remaining twenty-nine team members? Obviously twenty-nine.

So if there are 30 ways to choose the captain and, given that choice, there are 29 ways to choose the vice-captain, how many ways are there to pick the pair of them?

Is there a formula for that kind of problem? Yes. But you do not need to stuff you mind with more formulas that you do not feel confident about when to apply. The trick is to break the problem down into simple parts where the tools you understand perfectly do apply. This is not a topic for mechanical application of two formulas. It involves careful, logical simplification of an apparently complex problem into simpler problems.

 

[khansaheb]

How can I possibly do that if I do not understand the logic behind it? I was hoping you could guide me to some sources I can read and understand from.

Practice.... practice ..... practice. By solving several hundreds of problems - and keeping a snapshot of most of those in your mind.

 

[khansaheb]

I had requested

Please share your calculations....

And you responded:

a) 12
b) 36
c) 24

I used the concept of Permutation here.

I assume you are sharing the "numerical answers" - not calculation.

Please explain your "steps" and thought-process to arrive at those numbers - that will help you learn/remember the process for the next time around.

 

[pineapplewithmouse]

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.

In question a, does the option Girl1, Boy1, Girl2, Boy2, Girl3 is different from Girl1, Boy2, Girl2, Boy1, Girl3?
It changes the answer if so

 

[Steven G]

In question a, does the option Girl1, Boy1, Girl2, Boy2, Girl3 is different from Girl1, Boy2, Girl2, Boy1, Girl3?
It changes the answer if so

If you look at the two sets of 5 people do they look different to you?

If they weren't different, then there would be just one way to sit them, namely GBGBG

 

[pineapplewithmouse]

If you look at the two sets of 5 people do they look different to you?

If they weren't different, then there would be just one way to sit them, namely GBGBG

Of course they would, but you know, math questions aren't always logical, the classic example is John has in his car 208 watermelons and etc.
So maybe in this question they are identical, or maybe they only care about gender, and not about the fact that every person is different, I do know, that's why I asked.

 

[Steven G]

Of course they would, but you know, math questions aren't always logical, the classic example is John has in his car 208 watermelons and etc.
So maybe in this question they are identical, or maybe they only care about gender, and not about the fact that every person is different, I do know, that's why I asked.

..and during an exam, in my opinion, it is a fair question to ask your instructor if the boys are all identical or not. Don't be afraid to ask as it is a fair question and your grade depends on it.

 

[khansaheb]

..and during an exam, in my opinion, it is a fair question to ask your instructor if the boys are all identical or not. Don't be afraid to ask as it is a fair question and your grade depends on it.

In an exam, if there is any ambiguity, I would clearly state my assumption, e.g.,

I assume that each boy has been assigned different name (B1, B2, etc.), hence they are distinguishable from each other.

 

[Steven G]

In an exam, if there is any ambiguity, I would clearly state my assumption, e.g.,

I assume that each boy has been assigned different name (B1, B2, etc.), hence they are distinguishable from each other.

The problem with that is you will most probably changing the level of difficulty. Now if you made the problem easier and your work was correct, then you want the test maker to give you full credit. One attitude is you deserve full credit since the test maker was unclear. Sometimes you can lose that battle. Seriously, just ask for clarification.

 

[WillisRoy]

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.

Permutations and combinations are branches of mathematics that deal with the different ways in which objects can be arranged or selected.

Permutations involve arranging objects in a specific order, while combinations involve selecting objects without regard to order.

For example, if you have three different objects A, B, and C, the permutations of these objects in groups of two are AB, AC, BA, BC, CA, and CB, while the combinations of these objects in groups of two are AB, AC, and BC (since BA, CA, and CB are duplicates when order is not considered).

Now, let's tackle the specific questions you mentioned:

a) If the girls and boys sit alternately, we can first choose the arrangement of the girls (3!) and the boys (2!), and then alternate between them, giving a total of 3! x 2! x 2 = 24 arrangements.

b) If a girl sits at each end of the row, we can first choose the two girls for the end positions (3C2 = 3), and then arrange the remaining three children in the middle (3! = 6), giving a total of 3 x 6 = 18 arrangements.

c) If the girls sit together and the boys sit together, we can first arrange the girls among themselves (3! = 6) and then arrange the boys among themselves (2! = 2), giving a total of 6 x 2 = 12 arrangements.

 

[Steven G]

a) If the girls and boys sit alternately, we can first choose the arrangement of the girls (3!) and the boys (2!), and then alternate between them, giving a total of 3! x 2! x 2 = 24 arrangements.

There is just one way to sit the 5 and that is GBGBG. There aren't enough boys to have a boy 1st. BGBGG doesn't work.

 

[Steven G]

c) If the girls sit together and the boys sit together, we can first arrange the girls among themselves (3! = 6) and then arrange the boys among themselves (2! = 2), giving a total of 6 x 2 = 12 arrangements.

In this case you need to multiply your answer by 2 as you can have BBGGG or GGGBB