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Combinations. Source: Manhattan Flashcards

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Combinations. Source: Manhattan Flashcards

by sparkles3144 » Sun Jun 15, 2014 12:53 am
8 students have been chosen to play for PCU's inter-collegiate basketball team. If every person on the team has an equal chance of starting, what is the probability that both TOM and Alex will start?
(Assume 5 starting positions)

a) (6!/3!3!)/(8!/5!3!) b) (6*5*4)/(8*7*6*5*4) c) (6!/3!3!)/(8*7*6*5*4)

Answer is A

Can someone please explain?
I am lost.

Thanks!
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by [email protected] » Sun Jun 15, 2014 1:02 am
Hi sparkles3144,

In this question, the "order" of the players starting DOES NOT MATTER, so we're dealing with a combinatorics question. Since it also asks for the probability of Tom and Alex being in the starting 5 players, we need to do probability "math" too.

The combination formula is N!/K!(N-K)!

Let's start with the number of possible groups of 5 players that could start. There are 8 players and we need to choose 5.

That's 8c5....

8!/(5!)(3!)

Now we need the various combinations that satisfy what we're looking for....

Putting Tom and Alex on our team eliminates 2 of the "spots", which means that there are 3 spots remaining and we have 6 players to choose from.

That's 6c3

6!/(3!)(3!)

So, the probability of getting Tom and Alex on our team is...

[6!/(3!)(3!)] / [8!/(5!)(3!)]

This answer can be simplified to:

20/56 = 5/14

Although the given answer choices don't appear to have gone that far.

Final Answer: A

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by GMATGuruNY » Sun Jun 15, 2014 1:53 am
sparkles3144 wrote:8 students have been chosen to play for PCU's inter-collegiate basketball team. If every person on the team has an equal chance of starting, what is the probability that both TOM and Alex will start?
(Assume 5 starting positions)

a) (6!/3!3!)/(8!/5!3!) b) (6*5*4)/(8*7*6*5*4) c) (6!/3!3!)/(8*7*6*5*4)
Of the 8 students, 5 are to be chosen to start.
P(Tom is chosen) = 5/8.
Of the 7 remaining students, 4 are to be chosen to start.
P(Alex is chosen) = 4/7.
Since we want both events to happen, we MULTIPLY the fractions:
5/8 * 4/7 = 5/14.

A: ( 6!/3!3! )/ ( 8!/5!3! ) = (5*4)/(8*7) = 5/14.

The correct answer is A.
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