Question:

<p><strong>Step 1:</strong> Interpret 'differ by almost 10' as |a-b|≤10, where a and b are distinct natural numbers with 1≤a,b≤100.</p><p><strong>Step 2:</strong> Without loss of generality, assume a<b. We need: b-a≤10 and b-a≥1 (since they're different).</p><p><strong>Step 3:</strong> Use complementary counting: Total ways to select 2 different numbers from 1 to 100 is $\binom{100}{2}$.</p><p><strong>Step 4:</strong> Subtract pairs where b-a>10 (i.e., differ by more than 10). For b-a≥11, count pairs where 1≤a<b≤100 and b-a≥11.</p><p><strong>Step 5:</strong> If b-a≥11, then a≤89 and b≥12. Equivalently, if we set b=a+d where d≥11, then a ranges from 1 to 100-d. For d=11 to 99, we get $\sum_{d=11}^{99}(100-d) = \sum_{k=1}^{89}k = \binom{90}{2}$.</p><p><strong>Step 6:</strong> Alternatively, pairs with b-a≥11 correspond to choosing 2 numbers from {1,2,...,90} (if we relabel appropriately) which gives $\binom{90}{2}$.</p><p><strong>Step 7:</strong> Therefore, pairs differing by almost 10 = $\binom{100}{2} - \binom{90}{2}$</p><p>∴ Answer: $\binom{100}{2} - \binom{90}{2}$</p>