permutations (1 Viewer)

[purpletree]

Four girls and four boys are to sit in two horizontal rows of four. find the number of permutations if


A) Two boys want to sit in the front row, and a girl wants to sit in the back row

Attempt: 2! x 3! x 4! x 5!

So apparently my answer is wrong [5760]. Can someone please correct me?

 

[purpletree]

waaaa ): bump

 

[Carrotsticks]

Can you explain the logic behind your answer?

Perms + Combs is best learnt by you first giving us an answer (incorrect in this case), then us correcting what was wrong with the logic behind your answer.

 

[Shadowdude]

Four girls and four boys are to sit in two horizontal rows of four. find the number of permutations if


A) Two boys want to sit in the front row, and a girl wants to sit in the back row

Attempt: 2! x 3! x 4! x 5!

So apparently my answer is wrong [5760]. Can someone please correct me?

What's your working? How did you get 2!3!4!5!?

As I always say, do these logically:

1. Two boys want to sit in the front row, so place them there - choose two seats from the available four so they can sit there: C(4,2) ways
2. One girl wants to sit in the back row, so place her there - choose one seat from the available four: 4 ways
3. There are five people left, so place them wherever: 5! ways

Answer: 5!*4*C(4,2) = 2880.

Is that the right answer?

 

[purpletree]

Well, i said Just pick any two boys, make them one group in the first row. So now there are 3 groups at the front. we can arrange the boys, 2! and the three groups, so 3!

In the last row, there are 4 groups, so 4! . Besides the two boys at the front and the one girl at the back, everyone else can be arranged any way so that's 5!

in total we have 2! x 3! x 4! x 5!

 

[purpletree]

the answer says 5760 in terry lee

 

[Carrotsticks]

What's your working? How did you get 2!3!4!5!?

As I always say, do these logically:

1. Two boys want to sit in the front row, so place them there - choose two seats from the available four so they can sit there: C(4,2) ways
2. One girl wants to sit in the back row, so place her there - choose one seat from the available four: 4 ways
3. There are five people left, so place them wherever: 5! ways

Answer: 5!*4*C(4,2) = 2880.

Is that the right answer?

You forgot to permute the two boys.

Well, i said Just pick any two boys, make them one group in the first row. So now there are 3 groups at the front. we can arrange the boys, 2! and the three groups, so 3!

In the last row, there are 4 groups, so 4! . Besides the two boys at the front and the one girl at the back, everyone else can be arranged any way so that's 5!

in total we have 2! x 3! x 4! x 5!

When we get questions like this and they mention two boys or something like that, they usually mean 2 SPECIFIC boys ie: not any two boys as you said.

 

[Shadowdude]

Ah crud, forgot to treat the two boys in the front row as two possibilities.

2880 x 2 = 5760.

Revised:

1. Two boys want to sit in the front row, so place them there - choose two seats from the available four so they can sit there: C(4,2) ways
2. Place the first of the two boys in one of the seats... 2 ways
3. Place the remaining boy who wants to sit in the front row... 1 way
4. One girl wants to sit in the back row, so place her there - choose one seat from the available four: 4 ways
5. There are five people left, so place them wherever: 5! ways

Answer: 2*4*5!*C(4,2) = 5760.

There.

Well, i said Just pick any two boys, make them one group in the first row. So now there are 3 groups at the front. we can arrange the boys, 2! and the three groups, so 3!

In the last row, there are 4 groups, so 4! . Besides the two boys at the front and the one girl at the back, everyone else can be arranged any way so that's 5!

in total we have 2! x 3! x 4! x 5!

This kinda confuses me, to be honest. I kinda get it but... as I showed above, it can be done in a simpler, more straight forward way.

But if that works for you...

 

[purpletree]

You forgot to permute the two boys.



When we get questions like this and they mention two boys or something like that, they usually mean 2 SPECIFIC boys ie: not any two boys as you said.

woops. yep, that's what i meant. which was why i didnt do 4C2

 

[purpletree]

oh i get it now! thanks guys

 

[purpletree]

wait now im confused. why isnt this correct? :

1. there are four different groups in the first row: 4!
2. there are also four different groups in the second row: 4!
3. the remaining 5 people can rearrange themselves in anyway: 5!

4!x4!x5!

 

[Carrotsticks]

Well, i said Just pick any two boys, make them one group in the first row. So now there are 3 groups at the front. we can arrange the boys, 2! and the three groups, so 3!

In the last row, there are 4 groups, so 4! . Besides the two boys at the front and the one girl at the back, everyone else can be arranged any way so that's 5!

in total we have 2! x 3! x 4! x 5!

oh i get it now! thanks guys

As Shadowdude said, the best answer is the one that makes sense to you. Everybody thinks uniquely so quite often, we have original solutions for combinatoric problems.

Your logic (as different as it was to Shadowdude's) would have actually worked if you omitted the bolded part.

You double-counted the permutations of the groups.

 

[Carrotsticks]

wait now im confused. why isnt this correct? :

1. there are four different groups in the first row: 4!
2. there are also four different groups in the second row: 4!
3. the remaining 5 people can rearrange themselves in anyway: 5!

4!x4!x5!

Because you double-counted the arrangements of the people in the first and second row.

You had 4! meaning there are 4 people there to arrange. 2 of those are the boys meaning the other 2 are the 'leftover people'.

But you arranged them again when you did the 5! afterwards, so your answer will be too high.

 

[purpletree]

Because you double-counted the arrangements of the people in the first and second row.

You had 4! meaning there are 4 people there to arrange. 2 of those are the boys meaning the other 2 are the 'leftover people'.

But you arranged them again when you did the 5! afterwards, so your answer will be too high.

How would you have done it?

 

[Carrotsticks]

How would you have done it?

Same as Shadowdude:

P(4,2) x C(4,1) x 5!

 

[D94]

It's quite straightforward: if 2 boys want to sit in any of the front seats, they can be arranged P(4,2) ways, and likewise for the back row, 1 girl wants to sit there, so it's P(4,1). Now, there are 5 remaining seats for 5 people (doesn't matter whether boy or girl), so they can arranged P(5,5) or 5! ways. Multiply them together and you get 5760.

 

[Skeptyks]

It's quite straightforward: if 2 boys want to sit in any of the front seats, they can be arranged P(4,2) ways, and likewise for the back row, 1 girl wants to sit there, so it's P(4,1). Now, there are 5 remaining seats for 5 people (doesn't matter whether boy or girl), so they can arranged P(5,5) or 5! ways. Multiply them together and you get 5760.

Yeah I was wondering why they used a combination as opposed to a permutation.

 

[Shadowdude]

Yeah I was wondering why they used a combination as opposed to a permutation.

Same thing, but I feel more comfortable using combinations and then permuting them.

 

[Drongoski]

Attempt: 2! x 3! x 4! x 5!

So apparently my answer is wrong [correct answer: 5760].

You could have made it clearer. I got 5760 last night and thought it was the same as your wrong answer.

 

[ahdil33]

I got 5760 and thought I was wrong til I realised that was the answer you meant.

I just thought of it like this. Let's deal with the conditions first. We have one girl who has 4 possible seats she can sit in, in the back row

4...

Then we have the boys. The first boy can sit into another 4 possible seats, in the front row

4 X 4 ...

Next is the other boy, who can sit in 3 other seats left in the front row...

4 X 4 X 3

Now all restrictions are dealt with, and you have 5 people who left who don't care where they are. That means 5! ways of organising these five.

So, 4 X 4 X 3 X 5! = 5760