Establish the number of ways in which 7 different books can be placed on a bookshelf if 2 particular books must occupy the end positions and 3 of the remaining books are not to be placed together.
5! - 3! x 3!
5!-3! x 3! =84 and that is not the answer.
multiply it by 2
as you can permute the ennds in 2! ways
Can you draw it, or explain your steps?
(5! - 3! x 3!) x 2!
im thinking its more like 2*3*2*2*1*1*1 reasoning the first slot can only be filled with 2 books the 2nd slot can only be filled with 3 books cause out of 5 books the 3 books must occupy the 1st, 3rd and5th slots 3rd can only be filled with remaining 2 books, 4 is filled with remaining 2 books from the 3 and last remaining books fill the last 3 spots
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you fixed the ends with 2 books
remaining 5 books u can permute in 5! ways
yep.
but out of these 5! ways, there will be few arrangements, where the 3 books are together which u dont want
these form 3! x 3! ways
just subtract these from 5! to get the arrangements without these 3 books together
how do you know/where do you get 3 books are together is 3! x 3! ways?
good q
5 books
you can arrange them such that 3 books are always together in 3! x 3! ways
how ? thats ur q, correct ?
yes.
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say those are the 3 books u want always together
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