Question:

<p><strong>Step 1:</strong> Use Legendre's formula to find the highest power of prime <span class="math">p\</span> in <span class="math">n!\</span>:</p><p><span class="math">E_p(n!) = \left\lfloor \frac{n}{p} \right\rfloor + \left\lfloor \frac{n}{p^2} \right\rfloor + \left\lfloor \frac{n}{p^3} \right\rfloor + \ldots\</span></p><p><strong>Step 2:</strong> Calculate <span class="math">E_3(50!)\</span>:</p><p><span class="math">E_3(50!) = \left\lfloor \frac{50}{3} \right\rfloor + \left\lfloor \frac{50}{9} \right\rfloor + \left\lfloor \frac{50}{27} \right\rfloor + \left\lfloor \frac{50}{81} \right\rfloor + \ldots = 16 + 5 + 1 + 0 = 22\</span></p><p><strong>Step 3:</strong> Calculate <span class="math">E_3(40!)\</span>:</p><p><span class="math">E_3(40!) = \left\lfloor \frac{40}{3} \right\rfloor + \left\lfloor \frac{40}{9} \right\rfloor + \left\lfloor \frac{40}{27} \right\rfloor + \ldots = 13 + 4 + 1 + 0 = 18\</span></p><p><strong>Step 4:</strong> Calculate <span class="math">E_3(10!)\</span>:</p><p><span class="math">E_3(10!) = \left\lfloor \frac{10}{3} \right\rfloor + \left\lfloor \frac{10}{9} \right\rfloor + \left\lfloor \frac{10}{27} \right\rfloor + \ldots = 3 + 1 + 0 = 4\</span></p><p><strong>Step 5:</strong> For <span class="math">\binom{50}{10}\</span>:</p><p><span class="math">E_3\left(\binom{50}{10}\right) = E_3(50!) - E_3(10!) - E_3(40!) = 22 - 4 - 18 = 0\</span></p><p>∴ The highest power of 3 in <span class="math">\binom{50}{10}\</span> is <strong>0</strong>.</p>