If $z_1, z_2, z_3$ are complex numbers such that $|z_1|=|z_2|=|z_3|=1$ then $|z_1-z_2|^2+|z_2-z_3|^2+|z_3-z_1|^2$ cannot exceed:
Step-by-Step Solution
Key Concept: Legendre's formula for finding highest power of a prime dividing a factorial, combined with combinatorial counting under symmetry constraints.
For $E_2([125])$, we sum $\frac{[125]}{2} + \frac{[125]}{2^2} + \frac{[125]}{2^3} + \ldots = 62 + 31 + 15 + 7 + 3 + 1 = 119$. For $E_5([125])$, we calculate $\frac{[125]}{5} + \frac{[125]}{5^2} + \frac{[125]}{5^3} = 25 + 5 + 1 = 31$. The problem involves counting arrangements with constraints: for 4 items, Case 1 gives $1 \times ^3C_1 \times \frac{4}{2}$ and Case II gives 4 distinct arrangements. For 5 items with similar constraints, total arrangements equal 90, yielding $^nC_2 = 5050$.
Correct Answer: 2,3