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factorial - Why does 0! = 1? - Mathematics Stack Exchange The theorem that $\binom {n} {k} = \frac {n!} {k! (n-k)!}$ already assumes $0!$ is defined to be $1$ Otherwise this would be restricted to $0 <k < n$ A reason that we do define $0!$ to be $1$ is so that we can cover those edge cases with the same formula, instead of having to treat them separately We treat binomial coefficients like $\binom {5} {6}$ separately already; the theorem assumes
calculus - Nice proofs of $\zeta (4) = \frac {\pi^4} {90 . . . I know some nice ways to prove that $\\zeta(2) = \\sum_{n=1}^{\\infty} \\frac{1}{n^2} = \\pi^2 6$ For example, see Robin Chapman's list or the answers to the question "Different methods to compute $\\s
Find the inverse of a $4\times4$ matrix - Mathematics Stack Exchange Take a $4\times 8$ matrix whose first 4 columns are your matrix, and the last 4 columns are the identity Then row-reduce until you get the identity on the first 4 columns The matrix you get on the last four columns is the inverse of your matrix