(i) To have exactly one 'L', we choose 1 'L' from the 2 available 'L's and then choose 2 more letters from the remaining 6 letters (C, O, I, D, E, R). This is calculated as:

\(\binom{2}{1} \times \binom{6}{2} = 2 \times 15 = 15\)

(ii) Without restrictions, we consider three cases: no 'L', one 'L', and two 'L's.

• No 'L': Choose 3 letters from the 6 letters (C, O, I, D, E, R):

\(\binom{6}{3} = 20\)

• One 'L': Already calculated in part (i) as 15.
• Two 'L's: Choose 1 more letter from the 6 letters (C, O, I, D, E, R):

\(\binom{6}{1} = 6\)

\(Total selections = 20 + 15 + 6 = 41\)