(ii) The word TELEPHONE has 9 letters, with the letters T, L, P, H, N, and O available when there are no Es. This gives us 6 letters to choose from. The number of ways to choose 4 letters from these 6 is given by the combination formula:
\(\binom{6}{4} = 15\)
(iii) If there is exactly 1 E, we choose 1 E and then select 3 more letters from the remaining 6 letters (T, L, P, H, N, O). The number of ways to choose 3 letters from these 6 is:
\(\binom{6}{3} = 20\)
(iv) With no restrictions, we consider all possible selections of 4 letters from the 9 letters, accounting for the presence of Es. We calculate the number of ways for each scenario:
- No Es: \(\binom{6}{4} = 15\)
- 1 E: \(\binom{6}{3} = 20\)
- 2 Es: Choose 2 Es and 2 more letters from the remaining 6: \(\binom{6}{2} = 15\)
- 3 Es: Choose 3 Es and 1 more letter from the remaining 6: \(\binom{6}{1} = 6\)
Total number of ways: \(15 + 20 + 15 + 6 = 56\)
