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problems - Please discuss [#permalink] New post 17 Aug 2009, 21:33
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In a group of 8 semifinalists , all but 2 will advance to the final round. If in the final round only the top 3 will be awarded medals, then how many groups of medal winners are possible?

a) 20 b) 56 c) 120 d) 560 e) 720

The prob of rain on each of the 5 days is 1/6, except on the first day, when it is 2/5 and on the last day when it is 4/5. what is the probability that the rain will occur on atleast one of the five days?

a) 1/675 b) 5/72 c)5/27 d) 22/27 e) 67/72

coach Miller is filling out the starting line up for his indoor soccer team. There are 10 boys on the team, and he must assign 6 starters to the following positions : 1 goalkeeper , 2 on defense , 2 in mid feild , and 1 forward. Only 2 of the boys can play goal keeper and they cannot play any other positins. The other boys can play any other positions. How many different grouping are possible?

a) 60 b) 210 c) 2580 d) 3360 e) 151200

Note : All are taken from Princeton classroom book. I do not have the OA. Please discuss
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Re: problems - Please discuss [#permalink] New post 18 Aug 2009, 11:37
tkarthi4u wrote:
In a group of 8 semifinalists , all but 2 will advance to the final round. If in the final round only the top 3 will be awarded medals, then how many groups of medal winners are possible?

a) 20 b) 56 c) 120 d) 560 e) 720


SOL:
INCORRECT WAY:
6 out of 8 will advance to the final round. This can be done in 8C6 = 28 ways
Out of these 6, 3 will be awarded medals. This can be done in 6C3 = 20 ways
Total ways = 28 * 20 = 560
The reason this is incorrect is that we are counting duplicate groupings here. If a different combination of 6 semifinalists go to the next round but the SAME three semifinalists win the medals we are counting them as separate groupings; this is wrong!


CORRECT WAY:
A group of three medal winners can be selected from 8 semifinalists in 8C3 = 56 ways
ANS: B
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Last edited by samrus98 on 19 Aug 2009, 02:50, edited 1 time in total.
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Re: problems - Please discuss [#permalink] New post 18 Aug 2009, 11:45
tkarthi4u wrote:
The prob of rain on each of the 5 days is 1/6, except on the first day, when it is 2/5 and on the last day when it is 4/5. what is the probability that the rain will occur on atleast one of the five days?

a) 1/675 b) 5/72 c)5/27 d) 22/27 e) 67/72



SOL:
P(Rain on atleast one of the five days) = 1 - P(Rain on none of the five days)
P(Rain on none of the five days) = P(No rain on day 1) * P(No rain on day 2) * P(No rain on day 3) * P(No rain on day 4) * P(No rain on day 5)
= (1 - 2/5) * (1 - 1/6) * (1 - 1/6) * (1 - 1/6) * (1 - 4/5)
= 3/5 * 5/6 * 5/6 * 5/6 * 1/5
= 5/72

Thus P(Rain on atleast one of the five days) = 1 - 5/72 = 67/72

ANS: E
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Re: problems - Please discuss [#permalink] New post 18 Aug 2009, 11:57
tkarthi4u wrote:
Coach Miller is filling out the starting line up for his indoor soccer team. There are 10 boys on the team, and he must assign 6 starters to the following positions : 1 goalkeeper , 2 on defense , 2 in mid feild , and 1 forward. Only 2 of the boys can play goal keeper and they cannot play any other positins. The other boys can play any other positions. How many different grouping are possible?

a) 60 b) 210 c) 2580 d) 3360 e) 151200



SOL:
1 goalkeeper out of 2 can be selected in 2C1 ways
[strike]The vacant 5 positions can be filled by the remaining 8 boys in 8P5 = 6720 ways[/strike]

Since number of groupings is being asked, the above calculation is not right. This is a selection question rather than an arrangement question.

Now 2 on defense can selected in 8C2 ways, 2 in midfield in 6C2 ways and 1 in forward in 4C1 ways.

Total ways = 2C1 * 8C2 * 6C2 * 4C1 = 3360

ANS: D
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Last edited by samrus98 on 18 Aug 2009, 13:32, edited 1 time in total.
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Re: problems - Please discuss [#permalink] New post 18 Aug 2009, 13:01
tkarthi4u wrote:
In a group of 8 semifinalists , all but 2 will advance to the final round. If in the final round only the top 3 will be awarded medals, then how many groups of medal winners are possible?

a) 20 b) 56 c) 120 d) 560 e) 720

The prob of rain on each of the 5 days is 1/6, except on the first day, when it is 2/5 and on the last day when it is 4/5. what is the probability that the rain will occur on atleast one of the five days?

a) 1/675 b) 5/72 c)5/27 d) 22/27 e) 67/72

coach Miller is filling out the starting line up for his indoor soccer team. There are 10 boys on the team, and he must assign 6 starters to the following positions : 1 goalkeeper , 2 on defense , 2 in mid feild , and 1 forward. Only 2 of the boys can play goal keeper and they cannot play any other positins. The other boys can play any other positions. How many different grouping are possible?

a) 60 b) 210 c) 2580 d) 3360 e) 151200

Note : All are taken from Princeton classroom book. I do not have the OA. Please discuss


Q1,
suppose A B C get into top 3, and if consider A B C as a group other than individual winner, no orders for winner, that means top 1 A, top 2 B, top 3 C is the same as top 1 B, top 2 C and top 3 A,
the answer is 8C2*6C3= 560, choose D;

but if consider the orders among the winners,
then the answer should be 8C2*6P3=3360

Q2,
67/72

Q3,
goal keeper 2C1
defense 8C2
mid 6C2
forward 4C1
therefore, total formation 2C1*8C2*6C2*4C1=3360,
answer D
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Re: problems - Please discuss [#permalink] New post 18 Aug 2009, 18:54
Thank you.
1.OA was 2. They gave it as 8C3. I think we are right.

2 and 3 am not sure about OA.

One more doubts

P(Rain on atleast one of the five days) = 1 - P(Rain on none of the five days)

so if it is for atleast 2 days then i should do

P(Rain on atleast 2 of the five days) = 1 - [ P(Rain on none of the five days) + P(Rain on one of the five days) ]. Am i right. Please let me know.
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Re: problems - Please discuss [#permalink] New post 18 Aug 2009, 20:44
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Q1 should be

8C3=56

suppose there are 8 teams, T1, T2,...,T8,

and if T1 T2 are out, T3... T8 made into final; and T6 T7 T8 win the medal.
as you can see here,
[T1 T2][T3 T4 T5] lose
[T6 T7 T8] win

is the same as
[T3 T4][T1 T2 T5] lose
[T6 T7 T8] win

so, "560" contains quite a lot duplications. the correct answer is 8C3=56
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Re: problems - Please discuss [#permalink] New post 19 Aug 2009, 03:00
tkarthi4u wrote:
Thank you.
1.OA was 2. They gave it as 8C3. I think we are right.


8C3 is the correct answer. Have edited the solution above.


tkarthi4u wrote:
P(Rain on atleast 2 of the five days) = 1 - [ P(Rain on none of the five days) + P(Rain on one of the five days) ]. Am i right. Please let me know.


Yes, that is correct!
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Re: problems - Please discuss [#permalink] New post 22 Aug 2009, 02:24
i get it. Thanks samrus and flyingbunny.

Judges will select 5 finanlists 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 )1260

So here it is 7c3 * 3! = 210
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Re: problems - Please discuss [#permalink] New post 22 Aug 2009, 05:07
tkarthi4u wrote:
i get it. Thanks samrus and flyingbunny.

Judges will select 5 finanlists 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 )1260

So here it is 7c3 * 3! = 210

Yup. So basically we find all possible combinations of contestants in groups of 3. ie. 7C3 and then within each group 3 ribbons can be awarded in 3P3=3! ways.
So, total combinations = 7C3*3!

Good question involving both permutations, combinations and also the concept of duplication :)
+1 for the poster.
Re: problems - Please discuss   [#permalink] 22 Aug 2009, 05:07
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