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Mathematics 15 Online
OpenStudy (anonymous):

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.

OpenStudy (rsadhvika):

5! - 3! x 3!

OpenStudy (anonymous):

5!-3! x 3! =84 and that is not the answer.

OpenStudy (rsadhvika):

multiply it by 2

OpenStudy (rsadhvika):

as you can permute the ennds in 2! ways

OpenStudy (anonymous):

Can you draw it, or explain your steps?

OpenStudy (rsadhvika):

(5! - 3! x 3!) x 2!

OpenStudy (anonymous):

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

OpenStudy (rsadhvika):

|dw:1343113147136:dw|

OpenStudy (rsadhvika):

you fixed the ends with 2 books

OpenStudy (rsadhvika):

remaining 5 books u can permute in 5! ways

OpenStudy (anonymous):

yep.

OpenStudy (rsadhvika):

but out of these 5! ways, there will be few arrangements, where the 3 books are together which u dont want

OpenStudy (rsadhvika):

these form 3! x 3! ways

OpenStudy (rsadhvika):

just subtract these from 5! to get the arrangements without these 3 books together

OpenStudy (anonymous):

how do you know/where do you get 3 books are together is 3! x 3! ways?

OpenStudy (rsadhvika):

good q

OpenStudy (rsadhvika):

5 books

OpenStudy (rsadhvika):

you can arrange them such that 3 books are always together in 3! x 3! ways

OpenStudy (rsadhvika):

how ? thats ur q, correct ?

OpenStudy (anonymous):

yes.

OpenStudy (rsadhvika):

|dw:1343113608901:dw|

OpenStudy (rsadhvika):

|dw:1343113648052:dw|

OpenStudy (rsadhvika):

say those are the 3 books u want always together

OpenStudy (rsadhvika):

|dw:1343113694764:dw|

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