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June 2022 p52 q6
2763
(c) Four letters are selected from the 9 letters in the word CROCODILE. Find the number of selections in which the number of Cs is not the same as the number of Os.
(d) Find the number of ways in which the 9 letters in the word CROCODILE can be divided into three groups, each containing three letters, if the two Cs must be in different groups.
Solution
(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 \(\binom{5}{1} \times \binom{1}{1} \times 2! = 10\).
