To find the coefficient of $t^4$ in the expansion of ${\left(\frac{1-t^6}{1-t}\right)}^3$, we first simplify the expression inside the parentheses.

Step 1: Recognize that $\frac{1-t^6}{1-t}$ is a finite geometric series. The sum of a geometric series is given by:

$\frac{1-t^n}{1-t}=1+t+t^2+\dots +t^{n-1}$

For $n=6$, we have:

$\frac{1-t^6}{1-t}=1+t+t^2+t^3+t^4+t^5$

Step 2: Substitute this into the original expression:

${\left(\frac{1-t^6}{1-t}\right)}^3={(1+t+t^2+t^3+t^4+t^5)}^3$

Step 3: We need to find the coefficient of $t^4$ in the expansion of ${(1+t+t^2+t^3+t^4+t^5)}^3$. This is equivalent to finding the number of non-negative integer solutions to the equation:

$a_1+a_2+a_3=4$

where each $a_i$ can be 0, 1, 2, 3, 4, or 5 (since the exponents in the polynomial range from 0 to 5). However, since we are only looking for the coefficient of $t^4$, and 4 is less than 5, the upper limit of 5 does not restrict us here. So, we can ignore the upper limit for this case.

Step 4: The number of non-negative integer solutions to $a_1+a_2+a_3=4$ is given by the combination formula:

$\left(\frac{4+3-1}{3-1}\right)=\left(\frac{6}{2}\right)=15$

Therefore, the coefficient of $t^4$ is 15.

Final Answer: 15

Related Topics

This problem involves the expansion of a polynomial raised to a power, which is related to the Multinomial Theorem. The Multinomial Theorem generalizes the Binomial Theorem to more than two terms. It states that:

${(x_1+x_2+\dots +x_k)}^n=\sum \frac{n!}{n_1!n_2!\dots n_k!}{x}_1^{n_1}{x}_2^{n_2}\dots x_k^{n_k}$

where the sum is over all non-negative integers $n_1,n_2,\dots ,n_k$ such that $n_1+n_2+\dots +n_k=n$.

In this problem, we effectively used the concept of finding the number of solutions to an equation with constraints, which is a common combinatorial technique.

Formulae

Key formulae used:

1. Sum of a finite geometric series: $\frac{1-t^n}{1-t}=1+t+t^2+\dots +t^{n-1}$

2. Number of non-negative integer solutions to $x_1+x_2+\dots +x_r=n$ is $\left(\frac{n+r-1}{r-1}\right)$ or $\left(\frac{n+r-1}{n}\right)$.