Unformatted text preview: combinations of US members in the positions specified. We also have to place the other coun tries participants in the remaining 10 3 = 7 positions. This can be done in 7 4 , 2 , 1 = 7! 4! · 2! · 1! = 105 ways. So in total then we have 9 · 105 = 945 ways to position the participants. Problem 30 (seating delegates in a row) If the French and English delegates are to be seated next to each other, they can be can be placed in 2! ways. Then this pair constitutes a new “object” which we can place anywhere among the remaining eight people, i.e. there are 9! arraignments of the eight remaining people and the French and English pair. Thus we have 2 · 9! = 725760 possible combinations. Since in some of these the Russian and US delegates are next to each other, this number over counts the true number we are looking for by 2 · 28! = 161280 (the first two is for the number of arrangements of the French and English pair). Combining these two criterion we have 2 · (9!) 4 · (8!) = 564480(8!...
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This note was uploaded on 02/25/2011 for the course STAT 418 taught by Professor G.jogeshbabu during the Winter '08 term at Penn State.
 Winter '08
 G.JOGESHBABU
 Probability

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