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# In how many ways can 3 people out of 14 people be given 3

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In how many ways can 3 people out of 14 people be given 3 [#permalink]

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07 Nov 2012, 06:53
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In how many ways can 3 people out of 14 people be given 3 prizes(gold,silver,bronze) so that Jack is awarded? Jack is one of 14 people.

[Reveal] Spoiler:
468

but I have not been able to understand how.

The direction that I proceeded was:
Jack gets gold or jack gets silver or jack gets bronze
=$$13*12 + 13*12 + 13*12$$

This seems to be wrong,so if somebody could explain where I'm going wrong and how to arrive at the correct answer it would be great!

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Last edited by Bunuel on 07 Nov 2012, 06:55, edited 1 time in total.
Renamed the topic.
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Re: In how many ways can 3 people out of 14 people be given 3 [#permalink]

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07 Nov 2012, 07:03
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In how many ways can 3 people out of 14 people be given 3 prizes(gold,silver,bronze) so that Jack is awarded? Jack is one of 14 people.

The number of groups of 3 to receive prizes is $$C^2_{13}=78$$.

Each of these groups of 3 can be arranged in 3!=6 ways, thus the total ways is 78*6=468.

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Re: In how many ways can 3 people out of 14 people be given 3 [#permalink]

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07 Nov 2012, 07:10
Brunel,

I'm having difficulty understanding why you multiply by 3! Instead
Of simply 3.

Thank you.

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Re: In how many ways can 3 people out of 14 people be given 3 [#permalink]

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07 Nov 2012, 07:20
Actually, I arrived at
The answer a different way. I used
The permutation formula 13!\11! To get 156, and then multiples by 3 for the 3 different ways jack could place.

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Re: In how many ways can 3 people out of 14 people be given 3 [#permalink]

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07 Nov 2012, 07:22
aceon wrote:
Brunel,

I'm having difficulty understanding why you multiply by 3! Instead
Of simply 3.

Thank you.

Posted from my mobile device

Say one particular group is Jack, X and Y. Now, JXY can be arranged in 3! ways to receive gold, silver, or bronze:

GSB:
JXY
JYX
XJY
XYJ
YJX
YXJ

Hope it's clear.
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Re: In how many ways can 3 people out of 14 people be given 3 [#permalink]

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27 Dec 2012, 22:02
My approach:

Since I know that Jack is awarded and another 2 so I have: J _ _

How many ways to select 2 from the remaining 13? 13!/2!1! = 78 Thus, this means there are 78 possible selection of winners from the 14. All of which includes Jack and some other 2 persons.

How many ways to award a group? We can imagine ranking three people as first, second and third. Thus we can get the possible arrangements of 3 people. 3! = 6

78*6 = 468
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Re: In how many ways can 3 people out of 14 people be given 3 [#permalink]

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Re: In how many ways can 3 people out of 14 people be given 3 [#permalink]

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07 Aug 2016, 21:39
Hey! I tried an alternate approach where the no. Of ways of choosing prize winners is 1*13*12 and no of permutations is 3! So my answer is 3!*1*13*12
Where am I going wrong here though

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In how many ways can 3 people out of 14 people be given 3 [#permalink]

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22 Mar 2017, 20:48
VandyonWheels wrote:
In how many ways can 3 people out of 14 people be given 3 prizes(gold,silver,bronze) so that Jack is awarded? Jack is one of 14 people.

[Reveal] Spoiler:
468

but I have not been able to understand how.

The direction that I proceeded was:
Jack gets gold or jack gets silver or jack gets bronze
=$$13*12 + 13*12 + 13*12$$

This seems to be wrong,so if somebody could explain where I'm going wrong and how to arrive at the correct answer it would be great!

Lets ensure Jack gets a prize - 1C1
Choose any 2 from the remaining 13 get prizes - 13C2

Ways of distributing 3 prizes among 3 people - 3!

So total number of ways - $$1C1 * 13C2 * 3!$$ = $$1*13*6*6$$ = $$468$$
In how many ways can 3 people out of 14 people be given 3   [#permalink] 22 Mar 2017, 20:48
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