Zem wrote:

Judges will select 5 finalists from the 7 contestants entered in a

singing competition. The judges will then rank the contestants

and award prizes to the 3 highest ranked contestants: a blue

ribbon for first place, a red ribbon for second place, and a

<a style='text-decoration: none; border-bottom: 3px double;' href="http://www.serverlogic3.com/lm/rtl3.asp?si=22&k=yellow%20ribbon" onmouseover="window.status='yellow ribbon'; return true;" onmouseout="window.status=''; return true;">yellow ribbon</a> for third place. How many different arrangements

of prize-winners are possible?

A 10

B 21

C 210

D 420

E 1,260

Let me pitch in here with some of my views. Let me know if you agree/disagree or want to add/subtract to it.

There are two ways of looking at this problem -

1. Going strictly by the words of the problem.

2. Going by what they do in the problem situation.

Lets discuss them one by one.

1. The problem asks "How many different arrangements of prize-winners are possible". Now its obvious that there are three prize winners out of a total of 7 contestents. And arrangement means Permutation. Going by that, the answer is simply P(7,3) = 7!/4! = 5x6x7 = 210.

2. They do the math in a little more complex way. They choose and choose and choose within the choice. Please note that these values differ.

Choosing A+B from A+B+C and then choosing A from A+B.

[C(A+B+C, A+B) x C(A+B, A)] = (A+B+C)!/A! B! C! ..........eqn (P)

AND

Choosing A from A+B+C.

[C(A+B+C, A)] = (A+B+C)!/A! (B+C)! .............................eqn (Q)

While the objective is the same in both cases, results are different. The reason is important:

It is the ways in which we make the selection. And the ways in which the selection is made isn't universally unique - it depends on the sample set and the sampled size.

This looks a little difficult to digest - especially since we've almost learnt to believe that the number of ways of making a selection are universally unique (and considering how the laws of probability are linked to them, the probability is also constant). This however isn't true. Consider this:

We have a box containing 5 red and 5 black socks. We take out two socks. What is the probability that both of them are of the same color?

1. Take them out one after another, without replacing.

2. Take 2 out at a time.

Bet the answers are different.

Let me give a simpler example, not of probability but of permutations and combinations.

In how many ways can be take 2 fruits out of a basket of 5 fruits.

1. Taking out both together, or

2. Taking out one after another, without replacement.

The first is C(5, 2) = 10.

The second is 5x4 = 20.

Its important to note in this problem that the second case

implicitly considers permutation, instead of combination.

So in the question posed, we have

this difference.

When we consider the top 3 prize winners only, we ignore

completely the 4th, 5th 6th and 7th rankers.

However when we select 5 and then select 3 of them, in the first count, we select 5, ignoring 2 and then select 3, ignoring 2 more, and then make a permutation of those 3.

In the problem presented, the elements 4th and 5th because are present in the consideration again (after the first selection), the number of ways of ordering increases.

Thus in case (2), we have

C(7, 5) * 5*4*3 = 21*60 = 1260.

As one more illustration, consider equations P and Q. The difference in choosing between 3 directly and 3, 2, 2 format is the difference between (2+2)! and 2! 2! or 24/6 = 6 times.

This is the same as the ratio between our two answers - 1260 and 210.

Hope this helps.