Let the books be numbered \( 1 \) to \( 12 \). Now after the selection process, we label each book as \( A \) or \( B \) according to the following rule:- if a particular book is chosen, it is labelled \( A \) and if it is not chosen, it is labelled \( B \). Then we write the \( A \) \( B \) sequence. Note that each sequence corresponds to an unique selection of books. For example, the sequence \( ABABABABBBBB \) means that book \( 1 \) is chosen, book \(2\) isn't, book \( 3 \) is chosen, book \( 4 \) isn't, book \( 5 \) is chosen, book \( 6 \) isn't, book \( 7 \) is chosen, and books \( 8 \) to \( 12 \) aren't. Then our total number of acceptable permutations will be the number of ways of permuting \( 5 \) \( A \)s and \( 7 \) Bs such that no two \( A \)s are beside one another. To do this, place the 7 \(B\)s in gaps, like this \( _B_B_B_... \). Now there are \( 8 \) possible gaps and \( 5 \) gaps have to be filled by \( A \)s. This can be done in \( {8 \choose 5} \) ways.

can you explain the que, i can't understand what do u mean by "no 2 are consecutive"

Well.. the books are stacked side by side.. you're suppose to choose 5 that are not next to each other. I would solve this using complementary counting and then applying the principle of inclusion and exclusion. (I haven't tried it out yet.. so I'm not sure if it'll work)

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TopNewestLet the books be numbered \( 1 \) to \( 12 \). Now after the selection process, we label each book as \( A \) or \( B \) according to the following rule:- if a particular book is chosen, it is labelled \( A \) and if it is not chosen, it is labelled \( B \). Then we write the \( A \) \( B \) sequence. Note that each sequence corresponds to an unique selection of books. For example, the sequence \( ABABABABBBBB \) means that book \( 1 \) is chosen, book \(2\) isn't, book \( 3 \) is chosen, book \( 4 \) isn't, book \( 5 \) is chosen, book \( 6 \) isn't, book \( 7 \) is chosen, and books \( 8 \) to \( 12 \) aren't. Then our total number of acceptable permutations will be the number of ways of permuting \( 5 \) \( A \)s and \( 7 \) Bs such that no two \( A \)s are beside one another. To do this, place the 7 \(B\)s in gaps, like this \( _B_B_B_... \). Now there are \( 8 \) possible gaps and \( 5 \) gaps have to be filled by \( A \)s. This can be done in \( {8 \choose 5} \) ways.

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I got the answer as 56, i.e. 8C5.

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Well.. the books are stacked side by side.. you're suppose to choose 5 that are not next to each other. I would solve this using complementary counting and then applying the principle of inclusion and exclusion. (I haven't tried it out yet.. so I'm not sure if it'll work)

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can you explain the que, i can't understand what do u mean by "no 2 are consecutive"

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My answer is 41..please comment about my answer :))

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