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

Question

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 yellow ribbon for third place. How many different arrangements of prize-winners are possible?

A. 10
B. 21
C. 210
D. 420
E. 1,260

A. 10
B. 21
C. 210
D. 420
E. 1,260

Bunuel · Accepted Answer

Intermediary step of selecting 5 finalists first is just to confuse us: any group of 5 is equally likely to be selected, so we can skip this part and directly calculate ways to arrange 3 contestants out of 7:  C^3_7*3!=210 .

Answer: C.

kapslock · Answer

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.

  = (A+B+C)!/A! B! C! ..........eqn (P)

AND

Choosing A from A+B+C.
  = (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.

iwillcrackgmat · Answer

Simple Answer :

First Select 5 from 7 contestants and then select top 3 from those 5 =

7C5*5C3 = 42 * 5 = 210

mofasser08 · Answer

Bunuel, Would you go into details with your explanation, plz??

Bunuel · Answer

Ask yourself, why should # of arrangement be different for the case with intermediary step and for the case without it?

Or, HOW can # of arrangements of 3 people out 7 (with or without intermediary step) be more than  P^3_7  ( C^3_7*3!=210 )? I mean what different arrangement can there  possibly  be which is not counted in  P^3_7 ? Or which arrangement there won't be because of that intermediary step?

thinktank · Answer

Bunuel ,
 I went about this problem like this
 Since they are asking for total no of arrangements , i did 7*6*5 =210 . 7 ppl can win 1st place,6 ppl can win 2nd place and 5 ppl can win third. Is this approach correct or would this work only for this situation ?

Bunuel · Answer

Your approach is correct:
7 options for the first place;
6 options for the second place;
5 options for the third place.

Total 7*6*5=210.

SVaidyaraman · Answer

In my view, the answer cannot be  7P3 . The reason is, when we select 3 prize winners, we expect that all the three are finalists in the group of prize winners. But this is not the case if we include those who are not finalists while considering the arrangement of prize winners.

The possible number of arrangement of prize winners is 7C3 * 5P3.

Bunuel · Answer

OA is C (210).

You can try smaller numbers and list all cases to check that your approach won't give the correct answer.

SVaidyaraman · Answer

Then I think the OA is not correct. If the contestants are A, B, C, D, E, F and G and A, B, C, D, and E are finally selected how can F and G be among any list of prize winners?

Bunuel · Answer

OA is not wrong. Again, try to get the answer if it were 4 contestants with your approach and you'll see that your approach is not correct.

SVaidyaraman · Answer

Four Contestants : A B C D

Three are selected. This can be done in 4 C 3 ways - ABC, ABD, BCD, ACD

Consider 2 prizes

The prize winners from the above four combination of finalists can each be arranged in 3P2 ways

The answer is 4C3 * 3P2

Each of the 4C3 ways are independent of one another and so the arrangements have to be considered independently. It will be a repetition only within the same combination. So the ranking AB in ABC is independent of the same ranking AB in ABD. So we will have the prize winners AB in ABC and the prize winners AB in ABD and so on.

Your approach is obviously wrong because non finalists cannot be considered as prize winners.

Bunuel · Answer

That's not correct. Either you don't understand the question or the solutions.

Four Contestants : A B C D and 3 prizes. Possible cases:
ABC
ACB
BAC
BCA
CAB
CBA

ABD
ADB
BAD
BDA
DAB
DBA

ACD
ADC
CAD
CDA
DAC
DCA

BCD
BDC
CBD
CDB
DBC
DCB

Total of 24 cases.

My approach:  P^3_4=4!=24  --> correct.

ScottTargetTestPrep · Answer

Solution:

Since the finalists that do not make it to the top 3 ranks do not matter, we can ignore the selection of five finalists and calculate the number of ways to choose and order 3 contestants from a total of 7, which is 7P3 = 7!/(7 - 3)! = 7 x 6 x 5 = 210.

Answer: C

ak2121 · Answer

I am not very sure but i guess the way you got the answer is wrong, however the answer is right. We need to arrange the top 3 as well, so in this method, yoou will need to multiply 5c3 to 3! aswell.
However 5 finalists information, is redundant.
therefore 7c3*3!=210

iridescent995 · Answer

what is the meaning of  How many different arrangements of prize-winners are possible  ?

Doesn't it mean arrangement in a podium too?

Blue, red, yellow 
red, blue, yellow,
Y,R,B
...

shouldn't the answer be 3!*210??

E?

bumpbot · Answer

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

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