Question
In: Statistics and Probability
About finding the number of permutations Let there be n pairs of 2*n students : (1,...
About finding the number of permutations
Let there be n pairs of 2*n students : (1, 2) , (3, 4), (5, 6) ... (2n-1 , 2n). We want to find the number of arrangements of students which the pair are not adjacent. In other words, for (2*i) th student, the (2*i -1) th student should not be in his front or back.
For example, think of case of n=2. In this case, (1, 4, 3, 2) is not appropriate for this question because 4 and 3 are adjacent although they are pair. So we must count arrangements such as (1, 3, 2, 4)... etc for n=2
Let such total cases to be B_n. In this situation, how can we calculate the limit of (B_n)/ (2n!) ?
Solutions
orchestra answered 1 year agoRelated Solutions
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5
6
7
8
9
10
11
12
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4
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7
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5
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6
7
8
9
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4
3
4
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X
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1
3
2
3
3
7
4
13
5
13
6
1
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6
7
8
9
10
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9
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