Ask your own question, for FREE!
Mathematics 6 Online
OpenStudy (anonymous):

The total number of ways one can arrange 5 math books, 3 physics books, and 2 statistics books on a bookshelf, if no 2 books of the same subject can be placed next to each other, is...

OpenStudy (anonymous):

assume the books within each subject are distinct from each other

OpenStudy (anonymous):

hold on im looking at it

OpenStudy (anonymous):

10!

OpenStudy (anonymous):

5choose2 times 2factorial

OpenStudy (anonymous):

just 5 choose 2

OpenStudy (anonymous):

k

OpenStudy (anonymous):

you still there 2be?

OpenStudy (anonymous):

alright will u be on for the next 5 mins or so

OpenStudy (anonymous):

k thanks

OpenStudy (anonymous):

what does 5 permute one mean in words

OpenStudy (anonymous):

k im doing it by thinking of it as slots

OpenStudy (anonymous):

i get 1200 ways

OpenStudy (anonymous):

(5!*5!)/(3!*2!) = 1200

OpenStudy (phi):

Here's my shot at it

OpenStudy (phi):

OpenStudy (anonymous):

ok this is my reasoning

OpenStudy (anonymous):

the only way such a arrangement can be possible is if all the math books are spaced out so there are 2 possible ways this can happen, that the 10 slots start with a math book or end with a math book giving us a total of 5! * 2

OpenStudy (anonymous):

then , the remaining 5 slots in between the math books can be taken by the 5 other books in any order

OpenStudy (anonymous):

so to correct myself its 5!*5!*2

OpenStudy (anonymous):

28800

OpenStudy (anonymous):

hey wait phi is right

OpenStudy (anonymous):

i missed out on some other possibilities

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!