2013-10-23, 10:46 PM
nvm~ i'll go think about this for a sec
a)answer:
[(14 choose 7) * 14!/(7!7!)]/(21!/(7!7!7!)) =0.0295149....
1st term: choose the 7 spots out of the first 14 spots that the cordless phones go into. Since they are all the same there's no need to permute them.
2nd term: permute the other 14 phones in the open spots. There are 14 open spots... and two sets of 7 are identical phones.
division term: permuting 21 items with three sets of 7 identical items.
b)[3*(14 choose 7) * 14!/(7!7!)]/(21!/(7!7!7!))=0.08854....
multiply the top part by 3, so that any of the three phone types is entirely in the first 14 phones. This will leave the last 7 phones to be two of the three types.
c)[ (6!/(2!2!2!)) * (15!/(5!5!5!))]/ (21!/(7!7!7!))=0.1706656....
1st term: permute two of each phone type into the first six slots
2nd term: permute the rest of the phones
division term: all possible permutations.
hope these are right.... definitely sound right in my head.... but combinatorics is a very tricky (and fun) subject
a)answer:
[(14 choose 7) * 14!/(7!7!)]/(21!/(7!7!7!)) =0.0295149....
1st term: choose the 7 spots out of the first 14 spots that the cordless phones go into. Since they are all the same there's no need to permute them.
2nd term: permute the other 14 phones in the open spots. There are 14 open spots... and two sets of 7 are identical phones.
division term: permuting 21 items with three sets of 7 identical items.
b)[3*(14 choose 7) * 14!/(7!7!)]/(21!/(7!7!7!))=0.08854....
multiply the top part by 3, so that any of the three phone types is entirely in the first 14 phones. This will leave the last 7 phones to be two of the three types.
c)[ (6!/(2!2!2!)) * (15!/(5!5!5!))]/ (21!/(7!7!7!))=0.1706656....
1st term: permute two of each phone type into the first six slots
2nd term: permute the rest of the phones
division term: all possible permutations.
hope these are right.... definitely sound right in my head.... but combinatorics is a very tricky (and fun) subject