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can solve it this way?

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can solve it this way? [#permalink] New post 20 Nov 2016, 05:12
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hi,

see if i can sovle the below question like this?

six children will give presentation one after the other, in some order. Ruth's must come after Luke's(not necessarily immediately after Luke's).how many order are possible.?

i tried it like,

Ruth can take 5 position for presentation (except last poisition) = 5P1 = 5

Luke can take 5 position after ruth's presentation(cant take first position) = 5P1 = 5

rest can take remainng 4 position = 4P4= 24

multiplying all three = 5*5*24 = 600

am i right about my approach?
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Re: can solve it this way? [#permalink] New post 20 Nov 2016, 06:50
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Wrong explanation scroll down




There are 6 children say Eddy, Freddy, Teddy, Kiddie, Luke and Ruth.

Now Ruth must come after luke so we can combine them into one unit (Luke+Ruth). Now the problem becomes arranging Eddy, Freddy, Teddy, Kiddie , Luke+Ruth (5 objects).

You can arrange 5 objects in 5! ways or 120.

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Re: can solve it this way? [#permalink] New post 26 Dec 2016, 04:52
the question states not necessarily after
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Re: can solve it this way? [#permalink] New post 26 Dec 2016, 16:08
Expert's post
abhijith13 wrote:
the question states not necessarily after


Hey you are right. The new explanation.

Let us investigate Luke at different positions.

For example:
Luke at Position 1: Ruth can be at 5 different positions and other 4 can be at any position that available after Ruth.







Lukes PositionNumber of options for Ruths PositionNumber of ways to arrange the Remaining 4
Position 154!
Position 244!
Position 334!
Position 424!
Position 514!
Position 604!


Thus total number of ways of arrangement = \((5+4+3+2+1+0)*4!=360\).

Thanks for the correction.
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Re: can solve it this way?   [#permalink] 26 Dec 2016, 16:08
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