Eleven Fingers

Let the sum of the LHS be $N$ . Consider the operator ' $\pm$ ' be switches. So the number of switches is $9$ , and let the switch before $2$ be $s_2$ ; $3$ , $s_3$ ; $4$ , $s_4$ ; ... and $10$ , $s_{10}$ . Therefore, the largest $N$ is when all the switches are on ' $+$ ', $N_{Max} = 1+2+3$ $+4+5+6+7+8+9+10 = 55$ . We note that when we switch $s_n$ from on to off or ' $+$ ' to ' $-$ ', we minus $2n$ from $N_{Max}$ . For us to get $N=11 = 55 - 44$ $= 55 - 2(22)$ , we have to switch off $\sum n = 22$ . The acceptable ways are as follows and there are $\boxed{18}$ ways.

Let the sum of the numbers that are added on the RHS be $S_a$ and the sum of the numbers that are subtracted on the RHS be $S_s$ .

Note that $S_a=S_s+11$ . Since $S_a+S_s+11=66$ , then it follows that $S_a=33$ and $S_s+1=33\implies S_s=22$ .

It's easier to count $S_s$ , so we do so with a lot of listing and get $\boxed{18}$ .

Python 3.3:

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from itertools import product
operators = ('+', '-')
equation = '1{0}2{1}3{2}4{3}5{4}6{5}7{6}8{7}9{8}10 == 11'
count = 0
for combo in product(operators, repeat = 9):
    a, b, c, d, e, f, g, h, i = combo
    if eval(equation.format(a,b,c,d,e,f,g,h,i)):
        count += 1
print("Answer:", count)