An interesting problem of polynomials

The highest exponent possible is $1+2+ \cdots + 11 = 66$.

Now, to create the exponent $60$, you can only leave out the factors containing $(1,2,3),(2,4),(1,5)$ and $6$. Hence,

• $1+2+3 \Rightarrow$ gives coefficient $(-1)(-2)(-3) = 6$
• $2+4 \Rightarrow$ gives coefficient $(-2)(-4) = 8$
• $1+5 \Rightarrow$ gives coefficient $5$
• $6 \Rightarrow$ gives coefficient $-6$

Summing up gives $13-12 = \boxed{1}$.

Hint : $1+2+3 +...+11= \frac {11×12}{2} =66 $ so we must find how we can construct number $6=6+0=5+1=4+2=3+3=1+2+3$ and note that $3+3$ impossible.