8 people are sitting at a round table. 3 of them are chosen at random to give a presentation. What is the probability that the three chosen people were sitting in consecutive seats?
+118612 8 people are sitting at a round table. 3 of them are chosen at random to give a presentation. What is the probability that the three chosen people were sitting in consecutive seats?
the number of ways that 8 people can sit around a table is $${\mathtt{7}}{!} = {\mathtt{5\,040}}$$
the number of ways 3 distinct people can sit together around the table is $$(\left({\mathtt{6}}{\mathtt{\,-\,}}{\mathtt{1}}\right)){!}{\mathtt{\,\times\,}}{\mathtt{3}}{!} = {\mathtt{720}}$$
so I think the prob will be $${\frac{{\mathtt{1}}}{{\mathtt{7}}}}$$
$${\frac{{\mathtt{720}}}{{\mathtt{5\,040}}}} = {\frac{{\mathtt{1}}}{{\mathtt{7}}}} = {\mathtt{0.142\: \!857\: \!142\: \!857\: \!142\: \!9}}$$
+118612 8 people are sitting at a round table. 3 of them are chosen at random to give a presentation. What is the probability that the three chosen people were sitting in consecutive seats?
the number of ways that 8 people can sit around a table is $${\mathtt{7}}{!} = {\mathtt{5\,040}}$$
the number of ways 3 distinct people can sit together around the table is $$(\left({\mathtt{6}}{\mathtt{\,-\,}}{\mathtt{1}}\right)){!}{\mathtt{\,\times\,}}{\mathtt{3}}{!} = {\mathtt{720}}$$
so I think the prob will be $${\frac{{\mathtt{1}}}{{\mathtt{7}}}}$$
$${\frac{{\mathtt{720}}}{{\mathtt{5\,040}}}} = {\frac{{\mathtt{1}}}{{\mathtt{7}}}} = {\mathtt{0.142\: \!857\: \!142\: \!857\: \!142\: \!9}}$$
+128707 Melody is correct....let me expand on the second part of her answer, since it might not be clear how she arrived at that.....
Notice that if we "anchor" the 3 people in any three consecutive chairs in the circle.....there are 5! ways to arrange the other people in the remaining chairs.....but....there are also 3! ways to arrange the people occupying the consecutive seats.....!!!
So 5! x 3! = 120 x 6 = 720
