Probability question: In a bookshelf there are given 4 math, 3 english & 5 computer science books...?

cont...





in how many ways can you arrange the books if:


a.) there's no restriction


b.) books of the same subject are to be together


c.) the Math books are to be placed at the middle of English and computer science (books of the same subject may be together)


d.) computer science books are to be placed at the left side (books of the same subject may not be together)


e.) assuming that books of the same subject are of the same kind, how many ways can you arrange them?





4 MATH BOOKS


3 ENGLISH BOOKS


5 COMPUTER SCIENCE BOOKS





i badly need your help guys...





pls...thanks in advance...





god bless..(^_^)

Probability question: In a bookshelf there are given 4 math, 3 english %26amp; 5 computer science books...?
4M + 3E + 5C = 12 books in all





qa


no restriction


= 12P12


= 479001600





qb


same subject together


= 3P3 * [ 4P4 * 3P3 * 5P5 ]


= 6 * 24 * 6 * 120


= 103680





qc


each of them?


or all 4 M together in between of an E and a C?


i'll take the 2nd assumption. just because... it's way easier.





4M = 4P4


choosing 1E and 1C = 3C1 * 5C1


placing 1E1C = 2P2


arranging all the other = (5+3-2)P(5+#-2) = 6P6


placing "middled" 4M = 6+1





arranging c.)


= 4P4 * 3C1 * 5C1 * 2P2 * 6P6 *7


= 3628800





qd


computer science books are to be placed at the left side (books of the same subject may not be together)





not fully understanding, but since you typed "left side", so all 5 must be together. else, you'll get at least 1 of C in the "right side".





5C = 5P5


arranging Ms and Es = 1 (only way=MEMEMEM)


Ms = 4P4


Es = 3P3





arranging d.)


= 5P5 * 1 * 4P4 * 3P3


= 17280





qe


same subject are of the same kind


ok, but is this Q for (qa to qd), or just no restriction?





no restriction


= 15! / 4!5!6!


= 630630

morning glory