Arrange 4 Guests on One Side: 9C4*4!*9C3*3!*11!

[Crystal037]

Homework Statement

Eighteen guests have to be seated, half in each side of a long table. Four particular guests desire to sit on 1 particular side and the three others on the other side. Determine the no of ways in which sitting arrangement can be done

Relevant Equations

no of arrangements =nPr

4 guests want to be seated on one particular side.
So we choose 4 out of 9 seats on that particular side for them i.e 9C4. And they can arrange among themselves in 4! Ways.
So 9C4*4!
The we choose 3 out of 9 seats on the other side for the three people 9C3
They can arrange themselves in 3! Ways.
So 9C3*3!
Now the rests of the people can choose the 11 left seats and arrange themselves in 11! Ways
Therefore total no. of seating arrangements = 9C4*4! *9C3*3! * 11!
But thus isn't the correct answer.
Please tell me what is wrong with my approach

 

[BvU]

But thus isn't the correct answer.

If you've got it, post it!
Isn't half of eighteen equal to 9 any more ?

 

[Crystal037]

Oh sorry 9 would come in place of 8. Apart from that is there any error in my answer

 

[BvU]

But this isn't the correct answer.

How do you know ?
Do you know what form the answer has to be in (a number, an expression) ?

 

[Crystal037]

The hint given in the book is let the 4 particular guests be seated on side A and 3 on side B. So we are left with 11 guests out of which we choose 5 for side A and 6 for side B

 

[BvU]

Do you think your approach yields a different result than wat the book hint suggests ?

 

[Crystal037]

I don't know maybe

 

[BvU]

Compare $\ 11!\$ with ${11 \choose 6}\;6! \,5!$

 

[Crystal037]

So is my answer is correct?

 

[BvU]

How did you know the one with ${8\choose 5}\;5!\$ etc was not correct ?

 

[PeroK]

The hint given in the book is let the 4 particular guests be seated on side A and 3 on side B. So we are left with 11 guests out of which we choose 5 for side A and 6 for side B

I like your solution. However, you could look at it a different way. Suppose you really had this situation, how would you arrange the seating?

Hint: you could start with the 4 guests who want to sit on one side and, starting with the first of these, ask them to sit anywhere they want on that side.

 

[Crystal037]

How did you know the one with ${8\choose 5}\;5!\$ etc was not correct ?

Because my friend got the answer according to the book hint and the teacher told her that she's right. So, I thought that my approach was maybe wrong.

 

[Crystal037]

I like your solution. However, you could look at it a different way. Suppose you really had this situation, how would you arrange the seating?

Hint: you could start with the 4 guests who want to sit on one side and, starting with the first of these, ask them to sit anywhere they want on that side.

My first approach was this way only.

 

[PeroK]

My first approach was this way only.

I thought you chose the seats for them - rather than letting them sit where they wanted to?

 

[BvU]

Didn't your friend get the same number as you did ? Or was the answer not given as a number but as an expression ?
The comparison in #8 should result in: they are the same; so your answer is identical to the one of your friend.

 

[Crystal037]

Isn't choosing the seat is same as letting them sit where they want to?

 

[BvU]

It is -- for counting purposes

 

[Crystal037]

Didn't your friend get the same number as you did ? Or was the answer not given as a number but as an expression ?
The comparison in #8 should result in: they are the same; so your answer is identical to the one of your friend.

At that time I was not confident with my answer. So, I didn't compared the result. But now that I see it both approaches give the same result

 

[PeroK]

In any case, my solution would simply be:

$(9 \times 8 \times 7 \times 6)(9 \times 8 \times 7) \times 11!$

 

[BvU]

We had that already

 

[PeroK]

It's a lot simpler than using the book hint, which entails a lot of unnecessary binomial coefficients.

 

[Crystal037]

How the book hint entails a lot of unnecessary binomial coefficients

 

[PeroK]

How the book hint entails a lot of unnecessary binomial coefficients

Choosing 5 guests for one side of the table and 6 for the other. Just let them sit where they want!

The only thing you need to do is deal with the 4 and the 3 special guests first. And you can let them sit where they want as well. Minimal effort!

 

[BvU]

In short: @Crystal037 did well ! Kudos !

 

[Crystal037]

Thanks