C5. In the expression below, three of the + signs are changed into - signs so that the expression is equal to 100 :
$$
\begin{array}{l}
0+1+2+3+4+5+6+7+8+9+10 \\
+11+12+13+14+15+16+17+18+19+20
\end{array}
$$

In how many ways can this be done?
Solution
The given expression is equal to 210 .
Changing three signs from + to - will decrease the expression by double the sum of the three affected terms. So the sum of the three terms must be half of $210-100$, which is 55 .
The three terms must be distinct integers that are less than or equal to 20. The largest possible total for three such terms is $18+19+20$, which is 57 . To achieve a total of 55 , we need to decrease the sum of the three terms by 2 . This can be done by reducing the 18 by 2 or reducing the 19 by 2 (which is equivalent to taking one off each of the 18 and 19).
Hence, the expression can be changed to give a total of 100 in two ways - either by changing the signs on the 16,19 and 20 or on the 17,18 and 20 .