- #1
lesdavies123
- 16
- 0
Hi, this is the problem: Delegates from 10 countries, including Russia, France, England, and the United States, are to be seated in a row. How many different seating arrangements are possible if the French and English delegates are to be seated next to each other and the Russian and U.S. delegates are not to be next to each other.
So apparently the answer is 564 480 combinations, I come close to that, but not quite. Can anyone please correct my way of doing it which is as follows:
18 x (8!-14x(6!)) = 544 320 combinations
To justify my answer, the first 18 is for the 18 different combinations where England and France are sitting next to each other, then I tried to do the rest of the problem as if there were only 8 countries so if anyone sat next to anyone it would be 8! but - 14 (for all the different ways Russia and the US could be sitting next to each other) x 6! for all the different sitting patterns of the remaining countries! Obviously my way is wrong (unless a mistake in the answer key) so can anyone explain to me what I am doing wrong and what method would be better! Thank you very much in advance!
So apparently the answer is 564 480 combinations, I come close to that, but not quite. Can anyone please correct my way of doing it which is as follows:
18 x (8!-14x(6!)) = 544 320 combinations
To justify my answer, the first 18 is for the 18 different combinations where England and France are sitting next to each other, then I tried to do the rest of the problem as if there were only 8 countries so if anyone sat next to anyone it would be 8! but - 14 (for all the different ways Russia and the US could be sitting next to each other) x 6! for all the different sitting patterns of the remaining countries! Obviously my way is wrong (unless a mistake in the answer key) so can anyone explain to me what I am doing wrong and what method would be better! Thank you very much in advance!