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

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Re: Judges will select 5 finalists from the 7 contestants entered in a sin [#permalink]
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First Select 5 from 7 contestants and then select top 3 from those 5 =

7C5*5C3 = 42 * 5 = 210
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Re: Judges will select 5 finalists from the 7 contestants entered in a sin [#permalink]
Bunuel, Would you go into details with your explanation, plz??
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Re: Judges will select 5 finalists from the 7 contestants entered in a sin [#permalink]
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mofasser08 wrote:
Bunuel, Would you go into details with your explanation, plz??

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?
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Re: Judges will select 5 finalists from the 7 contestants entered in a sin [#permalink]
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Bunuel ,
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 ?
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Re: Judges will select 5 finalists from the 7 contestants entered in a sin [#permalink]
thinktank wrote:
Bunuel ,
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 ?

7 options for the first place;
6 options for the second place;
5 options for the third place.

Total 7*6*5=210.
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Re: Judges will select 5 finalists from the 7 contestants entered in a sin [#permalink]
christoph 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 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

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$$7C3 * 5P3.$$
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Re: Judges will select 5 finalists from the 7 contestants entered in a sin [#permalink]
SravnaTestPrep wrote:
christoph 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 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

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$$7C3 * 5P3.$$

OA is C (210).

You can try smaller numbers and list all cases to check that your approach won't give the correct answer.
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Re: Judges will select 5 finalists from the 7 contestants entered in a sin [#permalink]
Bunuel wrote:
SravnaTestPrep wrote:
christoph 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 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

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$$7C3 * 5P3.$$

OA is C (210).

You can try smaller numbers and list all cases to check that your approach won't give the correct 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?
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Re: Judges will select 5 finalists from the 7 contestants entered in a sin [#permalink]
SravnaTestPrep wrote:
Bunuel wrote:
SravnaTestPrep wrote:

OA is C (210).

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

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.
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Re: Judges will select 5 finalists from the 7 contestants entered in a sin [#permalink]
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Quote:
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.

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.
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Re: Judges will select 5 finalists from the 7 contestants entered in a sin [#permalink]
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SravnaTestPrep wrote:
Quote:
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.

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.

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
BDA
DAB
DBA

ACD
CDA
DAC
DCA

BCD
BDC
CBD
CDB
DBC
DCB

Total of 24 cases.

My approach: $$P^3_4=4!=24$$ --> correct.
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Re: Judges will select 5 finalists from the 7 contestants entered in a sin [#permalink]
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christoph 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 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

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.

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Re: Judges will select 5 finalists from the 7 contestants entered in a sin [#permalink]
iwillcrackgmat wrote:

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

7C5*5C3 = 42 * 5 = 210

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
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Re: Judges will select 5 finalists from the 7 contestants entered in a sin [#permalink]
Bunuel wrote:
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 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

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: $$C^3_7*3!=210$$.

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
...

E?
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Re: Judges will select 5 finalists from the 7 contestants entered in a sin [#permalink]
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Re: Judges will select 5 finalists from the 7 contestants entered in a sin [#permalink]
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