The unit digit of (3×1+4×2×1+5×3×2×1+ ... +20×18×17× ··· ×2×1) is

The expression is a sum of terms where each term can be simplified using factorial notation. The general term is $n \times (n-2)!$, which can be rewritten as $(n-1+1) \times (n-2)! = (n-1)! + (n-2)!$. Summing these terms from $n=3$ to $n=20$ results in a telescoping-like series: $(2!+1!) + (3!+2!) + (4!+3!) + \dots + (19!+18!)$. This simplifies to $2 \times (2! + 3! + \dots + 18!) + 1! + 19!$. To find the unit digit, we note that any factorial $n!$ for $n \ge 5$ ends in 0 because it contains factors of 2 and 5. Thus, we only need the unit digits of $2 \times (2! + 3! + 4!) + 1!$. Calculating these: $2! = 2$, $3! = 6$, and $4! = 24$. The sum is $2 \times (2 + 6 + 4) + 1 = 2 \times (12) + 1 = 24 + 1 = 25$. However, re-evaluating the specific series $3\times1 + 4\times2\times1 + 5\times3\times2\times1 \dots$ gives $3 + 8 + 30 + 144 + \dots$. The unit digits are $3 + 8 + 0 + 0 \dots = 11$, ending in 1. Wait, the expression is $3(1!) + 4(2!) + 5(3!) \dots + 20(18!)$. The unit digits are $3(1)=3$, $4(2)=8$, $5(6)=30 \to 0$. All subsequent terms $n \times (n-2)!$ for $n \ge 5$ contain a factor of 5 and an even number, so their unit digit is 0. Thus, the sum's unit digit is $3 + 8 = 11$, which ends in 1. Re-checking the options, if the first term is $3 \times 1!$, the sum is 11. If the sum is $2!+3!+4! \dots$, the unit digit is 2.