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Topic: permutation 2 (Read 424 times) |
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aks
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permutation 2
« on: Mar 15th, 2010, 12:13am » |
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2 Americans,2 British,2 Chinese and one each of duch,egiptial,french and german Persons are to be seated on the round table conference.
if the no of ways in which only american pairs are adjacent is equal to q*6!, Then find the Q
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| « Last Edit: Mar 15th, 2010, 9:31am by aks » |
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Grimbal
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Re: permutation 2
« Reply #1 on: Mar 15th, 2010, 3:14am » |
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Nobody from Finland?
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aks
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Re: permutation 2
« Reply #2 on: Mar 15th, 2010, 9:32am » |
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Quote:
Good Catch  : One is from france.
I have corrected the question now.
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rmsgrey
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Re: permutation 2
« Reply #3 on: Mar 16th, 2010, 11:41am » |
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Adapting my approach from the other problem, I get an answer of 640*6! = 460800 seating arrangments or 32*6! = 23040 if you only care about who's next to whom
Reasoning:
| hidden: | The As form a unit in one of two ways, leaving 9 units.
Each of the Bs needs a neighbour to their left who isn't a B. It matters whether that neighbour is a C, so there are three cases:
1) Neither neighbour is a C - 20 possibilities, for a total of 40 so far.
2) One neighbour is a C - 20 possibilities (two choices of which B gets paired with a neutral, two choices of which C gets paired, and 5 choices of neutral unit) for a total of 40 so far.
3) Both neighbours are Cs - 2 possibilities, for a total of 4 so far.
In any case, we're down to 7 units, two of which have a C on the left end.
The Cs also need non-C neighbours. Since cases 2) and 3) already have a non-C neighbour to the right of at least one C, it makes sense to go with right-neighbours:
1) We need to give both Cs neighbours, giving 20 cases, for a total of 800, with 5 units remaining.
2) Only one C needs a neighbour, so 5 cases, for a total of 200, with 6 units remaining.
3) Both Cs are already buddied up, so nothing needs doing, giving a total of 4, with 7 units remaining.
Arbitrarily deciding to put the As in the first and last seats, the number of arrangements of the remaining units varies by case:
1) 4! for a total of 800*4! or 160*5!
2) 5! for a total of 200*5!
3) 6! for a total of 4*6! or 24*5!
Combining the three cases gives 384*5! = 64*6! orders the people can be seated starting between the two Americans and going round clockwise or 32*6! orders if you don't care which direction you go in, or 640*6! possible seating arrangements if you're the person setting out the place cards. |
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