<p>A batsman scores exactly a century by hitting fours and sixes in twenty consecutive balls. In how many different ways can he do it if some balls may not yield runs and the order of boundaries and overboundaries are taken into account?</p>
Step-by-Step Solution
Key Concept: This is a problem of distributing 100 runs among 20 balls where each ball can contribute 0, 4, or 6 runs. You need to find non-negative integer solutions to 4f + 6s + 0n = 100 where f + s + n = 20, then count arrangements.
<p><strong>Step 1:</strong> Set up equations. Let f = fours, s = sixes, n = no-run balls.</p><p>Constraints: 4f + 6s + 0n = 100 and f + s + n = 20</p><p><strong>Step 2:</strong> From second equation: n = 20 - f - s. Substitute into first:</p><p>4f + 6s = 100, which simplifies to 2f + 3s = 50</p><p><strong>Step 3:</strong> Solve for non-negative integer solutions:</p><p>f = (50 - 3s)/2. For f to be a non-negative integer: 50 - 3s ≥ 0 and 50 - 3s must be even.</p><p>Since 50 is even and 3s must be even, s must be even. Let s = 0, 2, 4, 6, 8, 10, 12, 14, 16</p><p>Valid pairs (f, s, n):</p><p>• s=0: f=25, n=-5 ✗ (invalid)</p><p>• s=2: f=22, n=-4 ✗</p><p>• s=4: f=19, n=-3 ✗</p><p>• s=6: f=16, n=-2 ✗</p><p>• s=8: f=13, n=-1 ✗</p><p>• s=10: f=10, n=0 ✓</p><p>• s=12: f=7, n=1 ✓</p><p>• s=14: f=4, n=2 ✓</p><p>• s=16: f=1, n=3 ✓</p><p><strong>Step 4:</strong> For each valid solution, count arrangements (multinomial coefficient):</p><p>• (10,10,0): 20!/(10!×10!×0!) = 184,756</p><p>• (7,12,1): 20!/(7!×12!×1!) = 167,960</p><p>• (4,14,2): 20!/(4!×14!×2!) = 116,280</p><p>• (1,16,3): 20!/(1!×16!×3!) = 4,845</p><p><strong>Step 5:</strong> Total = 184,756 + 167,960 + 116,280 + 4,845 = 473,841</p><p>∴ <strong>Answer: 473,841 ways</strong></p>
Correct Answer: 473