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Probability - Baseball Team

Expert replies
by ntamhane » Tue Jul 26, 2011 7:34 pm
A baseball team consists of 20 players. 5 of whom are pitchers and 15 of whom are position players. If the order consists of 8 different position players and 1 pitcher, and if the pitcher always bats last in the order, which of the following expressions gives the number of possible different batting orders for this baseball team?

A. (15!)*(5)/8!
B. (15!)*(5)/7!
C. (15!)*(5!)/7!
D. 15!*5
E.20!

I go the answer [spoiler](15!)*5/8!*7! which is not a choice.[/spoiler] where am i going wrong
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Source: — Problem Solving |

by Frankenstein » Tue Jul 26, 2011 8:02 pm
Hi,
I guess you are just picking 8 players from 15. So, you used 15C8. But, here we have to arrange those 8 players. This can be done in 15P8, and the last position can be filled in 5 ways.
So, the number of ways should be (15P8)*5 = 15!*5/7!

Hence, B
Cheers!

Things are not what they appear to be... nor are they otherwise
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by krishnasty » Tue Jul 26, 2011 8:09 pm
I guess you are taking combination...
we have to 'arrange' players here...hence, use permutation
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Appreciation in thanks please!!
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