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manhatten problem

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manhatten problem

by nickhar130 » Thu Nov 04, 2010 1:18 pm
In how many ways can 6 people be seated at a round table if one of those seated cannot sit next to two of the other five?

a) 720
b) 120
c) 108
d) 84
e) 48

My weakness is definitely combination and permutation so please explain in detail. Greatly appreciated.
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by Taniuca » Thu Nov 04, 2010 4:58 pm
does someone knows how to calculate this?

so far this is what I get,
if all the persons could sit everywhere then 7!=720 options would be posibble. So answer a is eliminated.

If I had for instance A B C D E F, then ABC together could'nt be on the 1st or 2nd or third...up to six options* 2 because could be ( CBA) so that would give me 12 options that ABC or BCA cannot be sit. If I group ABC as a unit then I have 3 out of 4 options that D E F can be sit.
12 * 3C4= 48 so, the option e can be eliminated.

Since all the options that A B C D E F can be sitted is 5!=120 we can eliminate option b. I took this assuming that BC could still sit together. 120 minus the possible options were B could be located under A B C D E F , I get that
AC Cannot sit in certain possitions making a total of 12 not possible options to be sitted for instance, I show below if AC was together six possible options that times 2 because AC could be located as CA.
This gives me 120-12= 108 as the correct answer.

AC B D E F
B AC D E F
D B AC E F
D F B AC G
B D E F AC
AC D E F B
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