Question:

<p><strong>Step 1:</strong> Let the three distinct boxes contain x, y, z balls respectively, where x + y + z = 2t+1.</p><p><strong>Step 2:</strong> The constraint 'any two boxes together contain more than the third' translates to:</p><ul><li>x + y > z</li><li>y + z > x</li><li>z + x > y</li></ul><p><strong>Step 3:</strong> Since x + y + z = 2t+1, these become:</p><ul><li>x + y > (2t+1) - (x+y) ⟹ x + y > t + 0.5 ⟹ x + y ≥ t+1</li><li>Similarly: y + z ≥ t+1 and z + x ≥ t+1</li></ul><p><strong>Step 4:</strong> This means each box must contain ≤ t balls (since if any box had ≥ t+1, the sum of the other two would be ≤ t).</p><p><strong>Step 5:</strong> Total ways without restriction = {}^{2t+3}C_2 (stars and bars for distributing 2t+1 identical items into 3 distinct boxes).</p><p><strong>Step 6:</strong> Invalid cases: At least one box contains ≥ t+1 balls. If box i contains ≥ t+1 balls, we can substitute x_i' = x_i - (t+1) ≥ 0, giving x'_1 + x'_2 + x'_3 = t. This yields {}^{t+2}C_2 ways per box.</p><p><strong>Step 7:</strong> By inclusion-exclusion (no overlap for three boxes when total = 2t+1): Invalid = 3·{}^{t+2}C_2</p><p><strong>Step 8:</strong> Valid distributions = {}^{2t+3}C_2 - 3·{}^{t+2}C_2</p><p>∴ Answer: {}^{2t+3}C_2 - 3·{}^{t+2}C_2</p>