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can solve it this way?
[#permalink]
20 Nov 2016, 05:12
20 Nov 2016, 05:12
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?
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?
Re: can solve it this way?
[#permalink]
20 Nov 2016, 06:50
20 Nov 2016, 06:50
1
Expert Reply
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]
26 Dec 2016, 04:52
26 Dec 2016, 04:52
the question states not necessarily after
