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# Lets start with a simple one - Q - How many different

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Intern
Joined: 22 Jan 2006
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Lets start with a simple one - Q - How many different [#permalink]  30 Jan 2006, 20:59
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Lets start with a simple one -

Q - How many different arrangements of medal winners are possible out of 10 runners ?
A - 10!/7! ; in other words, 10*9*7 -> Right ?

Taking the above concept a step further,

Q - To fill a number of vacancies, an employer must hire 3 programmers from among 6 applicants, and 2 managers from among 4 applicants. What is the total number of ways in which she can make her selection ?

Which is the right answer ?

A1 : (6!/3!) * (4!/2!) = (6*5*4) * (4*3) = 1440

or

A2 : ((6*5*4)/(3*2*1)) * ((4*3)/(2*1)) = 120

Which answer is correct and why ?

How come the solution to the second question doesn't follow the first question's concept ?

What concept/logic does the second one use ?
GMAT Club Legend
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[#permalink]  30 Jan 2006, 21:20
The difference between the first question and second question lies in position.

In the first question, we're concerned with placing. Say we have ten runners, and the top three runners are Alvin, Simon and Sam.

But if we place the as:
Champ: Alvin
1st Runner up: Simon
2nd Runner up: Sam

It's different from the placing:
Champ:Sam
1st Runner Up:Alvin
2nd Runner up: Simon

Simiarly, this group is different from:
Champ: Simon
1st Runner up: Alvin
2nd Runner up: Sam

In all three scenarios, the top three runners are Alvin, Simon and Sam. But each are different in that we have different placings.

Now consider the second question.

"To fill a number of vacancies, an employer must hire 3 programmers from among 6 applicants, and 2 managers from among 4 applicants. What is the total number of ways in which she can make her selection ?"

If our 6 applicants are:
John, Mary, Simon, Alvin, Sam, Mark

and the eventual three that are hired are John, Mary and Simon, it doesn't matter if John was the first one hired, followed by Mary and finally Simon. Eventually, all we're concerned with is that John Mary and Simon are the new programmers in the company. (Note: The question clearly doesn't require placing)

Similarly, if the 4 applicants for the post of manager are Amy, Anna, Cathy and Ken, and two are hired: Amy and Anna. It doesn't matter if Amy was given the contract first, then Anna. We're only concerned that Amy and Anna are hired.

The fundamental I highlighted here is:

Question 1: Picking a number of items and ordering them (Permutation)

Question 2: Picking a number of items and order doesn't matter (Combination).

Let me know if it's not clear enough
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[#permalink]  30 Jan 2006, 21:24
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To answer your question:

Question 1: # of ways = 10P3 = 10!/7! = 10*9*8 = 720 ways

Question 2: # of ways = 6C3 * 4C2 = 20 * 6 = 120 ways
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[#permalink]  31 Jan 2006, 06:26
Thanks for the explanation. Could hear the school bells ringing !
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Re: Permutaions [#permalink]  14 Feb 2008, 10:18
dukapotamus wrote:
Lets start with a simple one -

Q - How many different arrangements of medal winners are possible out of 10 runners ?

Permuting 3 medals across 10 slots

10P3 = 10 * 9 * 8 = 720
_________________

You tried your best and you failed miserably. The lesson is 'never try'. -Homer Simpson

Re: Permutaions   [#permalink] 14 Feb 2008, 10:18
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# Lets start with a simple one - Q - How many different

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