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A talent contest has 8 contestants. Judges must award prizes

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A talent contest has 8 contestants. Judges must award prizes [#permalink] New post 30 May 2019, 11:26
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A talent contest has 8 contestants. Judges must award prizes for first, second, and third places, with no ties.

(a) In how many different ways can the judges award the 3 prizes?
(b) How many different groups of 3 people can get prizes?

[Reveal] Spoiler: OA
(a) 336 (b) 56


Math Review
Question: 10
Page: 297
Difficulty: medium

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Re: A talent contest has 8 contestants. Judges must award prizes [#permalink] New post 30 May 2019, 14:27
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(a) 8P3 =336
(b) 8C3 =56
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Re: A talent contest has 8 contestants. Judges must award prizes [#permalink] New post 31 May 2019, 04:43
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Carcass wrote:
A talent contest has 8 contestants. Judges must award prizes for first, second, and third places, with no ties.

(a) In how many different ways can the judges award the 3 prizes?
(b) How many different groups of 3 people can get prizes?

[Reveal] Spoiler: OA
(a) 336 (b) 56


Math Review
Question: 10
Page: 297
Difficulty: medium


(a) In how many different ways can the judges award the 3 prizes?

Take the task of awarding prizes and break it into stages.

Stage 1: Select a contestant to receive the FIRST prize
There are 8 contestants to choose from.
So, we can complete stage 1 in 8 ways

Stage 2: Select a contestant to receive the SECOND prize
There are 7 contestants REMAINING to choose from.
So, we can complete stage 2 in 7 ways

Stage 3: Select a contestant to receive the THIRD prize
There are 6 contestants REMAINING to choose from.
So, we can complete stage 3 in 6 ways


By the Fundamental Counting Principle (FCP), we can complete all 3 stages (and thus award the 3 prizes) in (8)(7)(6) ways (= 336 ways)

Answer: 336

-------------------------------------------------

(b) How many different groups of 3 people can get prizes?

In this question, the order in which we select the people doesn't matter.
That is, we're just selecting 3 people.
We can select 3 people from 8 people in 8C3 ways
8C3 = (8)(7)(6)/(3)(2)(1) = 56

Answer: 56

ASIDE: Here's a video on calculating combinations (like 8C3) in your head:

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Re: A talent contest has 8 contestants. Judges must award prizes [#permalink] New post 16 Feb 2020, 18:17
a) For this question, we have given that we need to choose the first 3 prizes without the repetition. So the formula is n!/(n-r)!

Where n is the number of things to choose from and we choose r of them, no repetitions, order matters.

So, lets apply it here. 8!/(8-3)!=8!/5!=8*7*6=336


b) How many different groups of 3 ppl can be selected?

The general formula for the number of combinations of r things from n different things is : n!/(n-r)!r!

So, lets apply it here : 8!/5!3!=(8*7*6)/(1*2*3)=56
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Re: A talent contest has 8 contestants. Judges must award prizes   [#permalink] 16 Feb 2020, 18:17
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