Both TOM and Alex will start

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Both TOM and Alex will start

by srinivasarajui » Tue Apr 06, 2010 3:12 am
What is the correct expression for this problem?

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)


Can any one explain me, how to get the answer.
The answer specified is a

Source : https://www.manhattangmat.com/flashcards.cfm - WT
Srinu
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by neoreaves » Tue Apr 06, 2010 3:30 am
I am not sure if i am right but is

a) (5!/3!3!)/(8!/5!3!) instead of (6!/3!3!)/(8!/5!3!)

5! IN PLACE OF 6! ?

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by indiantiger » Tue Apr 06, 2010 3:33 am
Here is how I would approach it

Given : total 8 students
team can be made of 5 students :

total possibilities = 8C5 = 8!/5!3!

Now Tom and Alex always need to be on the team so that leaves only 3 spots on the team and to be filled out by 6 people
(8 - tom - alex)
total ways in which a team can be formed now: 6C3 = 6!/3!3!

Probability = # favorable outcomes/total outcomes
= (6!/3!3!)/(8!/5!3!) (A)

I hope this helps.

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by srinivasarajui » Tue Apr 06, 2010 3:40 am
5! IN PLACE OF 6! ?
Yes you are right.

I got that, lot of thanks
Srinu