Permutations - placement in a line

Just having a bit of a problem with a particular permutation question.

Q - Four girls (Anne, Pela, Jeanette and Philippa) and three boys (Tony , Jason, and Bill) arrange themselves in a line at the checkout. If the arrangements are random, find the probability that Anne and Pela will be served before tony

My attempt.

5 possible situations, two girls always on the right side of the line, tony always on the left side of the line.

x x x x x | x x permutations 5P5 * 2!

x x x x | x x x permutations 5P4 * 3!

x x x | x x x x permutations 5P3 * 4!

x x | x x x x x permutations 5P2 * 5!

x | x x x x x x permutations 5P1 * 6!

add them up and put it over 7! and it comes out as 5/3 obviously wrong lol :S.

Re: Permutations - placement in a line

Quote:

Originally Posted by

**tomdoml** Just having a bit of a problem with a particular permutation question.

Q - Four girls (Anne, Pela, Jeanette and Philippa) and three boys (Tony , Jason, and Bill) arrange themselves in a line at the checkout. If the arrangements are random, find the probability that Anne and Pela will be served before tony

My attempt.

5 possible situations, two girls always on the right side of the line, tony always on the left side of the line.

x x x x x | x x permutations 5P5 * 2!

x x x x | x x x permutations 5P4 * 3!

x x x | x x x x permutations 5P3 * 4!

x x | x x x x x permutations 5P2 * 5!

x | x x x x x x permutations 5P1 * 6!

add them up and put it over 7! and it comes out as 5/3 obviously wrong lol :S.

You should consider how many ways there are to have the orders

APT ..... in other words Anne before Pela before Tony

PAT ..... which is Pela before Anne before Tony.

Those are the 2 possible orderings of the 3.

Now to find the number of ways we have APT in the 7,

we only need select 3 positions of the 7,

so we count the number of ways we can do that.

A before P before T occurs in $\displaystyle \binom{7}{3}$ ways.

The same calculation counts the number of ways Pela is before Anne who is before Tony.

Having counted the number of ways we can have APT,

there are 4! ways to arrange the remaining 4 people each time.

This will help to find the numerator of the probability fraction.

Needless to say, there are sure to be quicker ways.

For example, you may ask...

Is it just as likely that Tony will be in front of Pela and Anne,

in between Pela and Anne,

or behind Pela and Anne.

Re: Permutations - placement in a line

Hello, tomdoml!

I worked this out the "long way".

Upon seeing the answer, a simpler solution occurred to me.

Quote:

Four girls (Anne, Pela, Jeanette, Philippa) and three boys (Tony, Jason, Bill)

arrange themselves in a line at the checkout. .If the arrangements are random,

find the probability that Anne and Pela will be served before Tony.

Consider the order of $\displaystyle A$, $\displaystyle P$ and $\displaystyle T.$

There are $\displaystyle 3! = 6$ orderings:$\displaystyle APT, AT\!P, P\!AT, PT\!A, T\!AP, T\!P\!A$

In two orderings $\displaystyle \{APT, P\!AT\},\:A$ and $\displaystyle P$ precede $\displaystyle T.$

Therefore: .$\displaystyle P(A\text{ and }P\text{ before }T) \:=\:\frac{2}{3}$

[The other four people can be arranged in 4! ways, but *we don't care!*]

.

Re: Permutations - placement in a line

Another way to look at it...

The queue has 7 positions.

Suppose Tony picks the last position. There is a 1/7 chance of this

[T] [] [] [] [] [] []

Pela and Anne will certainly be ahead of him

Suppose Tony picks the 2nd last position. There is a 1/7 chance of this.

[] [T] [] [] [] [] []

The first girl has a 5/6 chance of being in front of him

and the other girl has a 4/5 chance

(since there will be 4 remaining positions in front of Tony to choose from)

Suppose Tony picks the next position up, again a 1/7 chance.

[] [] [T] [] [] [] []

The first girl has a 4/6 chance of being in front

and the 2nd girl has a 3/5 chance.

Therefore, the probability of Tony being behind the girls is

$\displaystyle \frac{1}{7}\left(\frac{6}{6}\;\frac{5}{5}+\frac{5} {6}\;\frac{4}{5}+\frac{4}{6}\;\frac{3}{5}+\frac{3} {6}\;\frac{2}{5}+\frac{2}{6}\;\frac{1}{5}\right)$

$\displaystyle =\frac{1}{7}\left(\frac{70}{30}\right)$

Now, if there were 100 people in the queue,

it helps to realise that the size of the queue is irrelevant !

Finally, Soroban may have meant to have written 2/6 rather than 2/3

If you imagine a queue of "any" length, say 20 people.

Tony, Pela and Anne are in the queue.

Then they could be in any 3 positions.

If we leave all the other people standing where they are,

then we can arrange Tony, Pela and Anne among each others 3 positions.

In 2 of these 6 arrangements, Tony will be behind the girls.

If Tony, Pela and Anne are in 3 different positions, the exact same logic applies,

so no matter which 3 positions those 3 people occupy,

there are always 2 chances in 6 that Tony will be behind the girls.