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Explanation:

T1 = 1 × 1! = (2 - 1) × 1! = 2 × 1! - 1 × 1! = 2! - 1!

T2 = 2 × 2! = (3 - 1) × 2! = 3 × 2! - 1 × 2! = 3! - 2!

T3 = 4! - 3!

... and so on.

∴ Let S = 1 × 1! + 2 × 2! + 3 × 3! + 4 × 4! + ⋯ 10 × 10!

⇒ S = (2! - 1!) + (3! - 2!) + (4! - 3!) + ... + (10! - 9!) + (11! - 10!)

⇒ S = 11! - 1! = 11! - 1

Hence, option (b).

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