(c) Consider the word CROCODILE, which contains the letters C, R, O, C, O, D, I, L, E. We need to select 4 letters such that the number of Cs is not the same as the number of Os.
Possible scenarios:
- 2 Cs and 0 Os: Choose 2 more letters from the remaining 5 letters (R, O, D, I, L, E), which is \(\binom{5}{2} = 10\).
- 0 Cs and 2 Os: Choose 2 more letters from the remaining 5 letters (C, R, D, I, L, E), which is \(\binom{5}{2} = 10\).
- 1 C and 0 Os: Choose 3 more letters from the remaining 5 letters (R, O, D, I, L, E), which is \(\binom{5}{3} = 10\).
- 0 Cs and 1 O: Choose 3 more letters from the remaining 5 letters (C, R, D, I, L, E), which is \(\binom{5}{3} = 10\).
- 1 C and 1 O: Choose 2 more letters from the remaining 5 letters (R, D, I, L, E), which is \(\binom{5}{2} = 10\).
\(Total = 10 + 10 + 10 + 10 + 10 = 50.\)
(d) We need to divide the 9 letters into three groups of three, ensuring the two Cs are in different groups.
- Both Os in a group with a C: Choose 1 more letter from the remaining 6 letters, which is \(\binom{5}{2} = 10\).
- Both Os in a group without a C: Choose 1 more letter from the remaining 6 letters, which is \(\binom{5}{2} \times \binom{3}{2} = 30\).
- One O in a C group, one not: Choose 1 more letter from the remaining 6 letters, which is \(\binom{5}{1} \times \binom{3}{2} = 30\).
- One O with each C: Choose 1 more letter from the remaining 6 letters, which is \(\binom{5}{1} \times \binom{1}{1} \times 2! = 10\).
\(Total = 10 + 30 + 30 + 10 = 80.\)
