clock60 wrote:govind_raj_76 wrote:A baseball team consists of 20 players, 5 of whom are pitchers and 15 of whom are position players. If the batting order consists of 8 different positions and 1 pitcher, and if the pitcher always bats in the last order, which of the following different batting orders for this baseball team ??
A) [(15 !) X (5)] / 8
A) [(15 !) X (5)] / 7 !
C) [(15 !) X (5 ! )] / 7 !
D) (15 !) X (5)
E) 20 !
not sure that i know exactly what is batting order, but my try
8 different players from 15 can be selected 15C8 ways and inside this group they have 8! permutations
1 pitcher fron 5 can be selected 5C1 ways, so
15C8*8!*5C1=(15!*5)/7!
i got B
Good solution!
Another way we could attack it is to use the permutations formula (which you did, albeit unknowingly).
When we're arranging k items out of a total pool of n items, the number of possible arrangements is:
nPk = n!/(n-k)!
Looking at the position players, we have 8 to arrange and 15 available, so:
15P8 = 15!(15-8)! = 15!/7!
Then we have 5 possible pitchers for the 9th spot, so we multiply by 5 to get:
15!/7! * 5
= (15! * 5)/7!
* * *
Your solution worked as well, since what you did was derive the permutations formula from the combinations formula.
nCk = n!/k!(n-k)!
The only difference between the two formulae is the extra k! in the bottom of the combinations formula. Since you multiplied by 8!, you cancelled out the k! in the bottom of the fraction and ended up with the right answer.
