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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59 randomly selected students were asked the number of pairs of shoes they have. Let X...
59 randomly selected students were asked the number of pairs of
shoes they have. Let X represent the number of pairs of shoes. The
results are as follows:
Round all your answers to 4 decimal places where
possible.
The mean is:
The median is:
The sample standard deviation is:
The first quartile is:
The third quartile is:
What percent of the respondents have at least 12 pairs of Shoes?
%
53% of all respondents have fewer than how many pairs...
64 randomly selected students were asked the number of pairs of shoes they have. Let X...
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shoes they have. Let X represent the number of pairs of shoes. The
results are as follows:
# of Pairs of
Shoes
4
5
6
7
8
9
10
11
12
13
14
15
Frequency
4
4
3
7
2
5
10
5
5
8
6
5
Round all your answers to 4 decimal places where
possible.
The mean is:
The median is:
The sample standard deviation is:
The...
57 randomly selected students were asked the number of pairs of shoes they have. Let X...
57 randomly selected students were asked the number of pairs of
shoes they have. Let X represent the number of pairs of shoes. The
results are as follows:
# of Pairs of Shoes
4
5
6
7
8
9
10
11
Frequency
7
8
14
4
3
4
7
10
Round all your answers to 4 decimal places where
possible.
The mean is: __________________
The median is: __________________
The sample standard deviation is: __________________
The first quartile is: __________________
The...
Forty randomly selected students were asked the number of pairs of sneakers they owned. Let X...
Forty randomly selected students were asked the number of pairs
of sneakers they owned. Let X = the number of pairs of
sneakers owned. The results are as follows.
X
Frequency
1
3
2
3
3
7
4
13
5
13
6
1
Find: Sample Mean, Standard Deviation, Relative Frequency and
Cumulative Relative Frequency, First and Third Quartiles, What
percent of the students owned at least five pairs? (Round your
answer to one decimal place.), 40th Percentile and 90th
Percentile.
57 randomly selected students were asked the number of pairs of shoes they have. Let X...
57 randomly selected students were asked the number of pairs of
shoes they have. Let X represent the number of pairs of shoes. The
results are as follows:
# of Pairs of
Shoes
4
5
6
7
8
9
10
11
Frequency
7
9
7
5
9
3
8
9
Round all your answers to 4 decimal places where
possible.
The mean is:
The median is:
The sample standard deviation is: _______
The first quartile is: _______
The third quartile...
Module/Week 2 ASSIGNMENT (INPUT/OUTPUT) The number of permutations of a set of n items taken r...
Module/Week 2 ASSIGNMENT (INPUT/OUTPUT)
The number of permutations of a set of n items taken r at a time is
given by the following
formulan!/r !(n- r )!: where n! is the factorial of n, r! is the
factorial of r, and (n-r)! is the
factorial of the result of n-r. The factorial of a number n can be
solved using the following
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fixed case k=3.
(i) Write clearly the statement (**) we wish to prove. Be sure
your statement includes the phrase “for all n” .
(ii) State explicitly the assumption in (**) we will thus
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(iii) Now recall that to prove by induction means to show that
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singleton blocks. And let A(n) be the number of all partitions of
[n] with at least one singleton block. Prove that for all n ≥ 1,
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argument.
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