1.
To compute the product of the two permutations π and t, we can simply apply the permutations one after the other. This means that we first apply the permutation π to the elements, and then apply the permutation t to the resulting elements.

The permutation π takes the elements a, b, c, d, e and rearranges them as follows:

a -> c
b -> b
c -> d
d -> e
e -> a

The permutation t takes the elements and rearranges them as follows:

a -> a
b -> b
c -> d
d -> e
e -> c

Applying the permutations π and t in sequence means that we first apply π to the elements, and then apply t to the resulting elements.

a -> c -> d
b -> b -> b
c -> d -> e
d -> e -> c
e -> a -> a

so, πt = [a b c d e; d b e c a]

2.
To find the inverse of a permutation, we need to find a permutation that, when composed with the original permutation, results in the identity permutation. The identity permutation is the permutation that leaves all elements unchanged, which can be represented as:

[a b c d e; a b c d e]

To find the inverse of π, we can simply apply the permutation π to each element in the identity permutation. For example, to find the inverse of the first row of π, we would do:

[π(a) π(b) π(c) π(d) π(e); a b c d e]

Plugging in the values from π, we get:

[c b d e a; a b c d e]

So the inverse of π is:

π^(-1) = [c b d e a; a b c d e] = [a b c d e; e b a c d]

To verify that this is indeed the inverse of π, we can check that π * π^(-1) = [a b c d e; a b c d e], which is the identity permutation.

[a b c d e; c b d e a] * [a b c d e; e b a c d] = [a b c d e; a b c d e]

3.
To compute the entropy of a random variable, we need to compute the probability of each possible outcome and then apply the formula for entropy.

In this case, the possible outcomes for the random variable X are 1, 2, 3, 4, and 5, each with equal probability of 1/5. Therefore, the entropy of X is:

H[X] = – ∑ p(x) * log(p(x))
= – (1/5) * log(1/5) – (1/5) * log(1/5) – (1/5) * log(1/5) – (1/5) * log(1/5) – (1/5) * log(1/5)
= -5 * (1/5) * log(1/5)
= -5 * (-0.6931)
= 3.4655

PS: The log in the calculation is base 10


Sorry, you confused me a bit. I don’t know what you’re talking about. I have not contacted you through any other means, only reply you here. 


I don\’t know how many times you sent the question, but I only answered this one, and I didn\’t send you an em-ail. It is possible that the teacher who answered another question sent you the em-ail.

Reminder
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