Question

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.

Answer 1

50

Answer 2

The word CROCODILE has 9 letters: C, R, O, C, O, D, I, L, E. The distinct letters and their frequencies are: C: 2 R: 1 O: 2 D: 1 I: 1 L: 1 E: 1 There are 7 distinct letter types: C, O (frequency 2), and R, D, I, L, E (frequency 1). Let the set of single letters be $S = \{R, D, I, L, E\}$, with $|S|=5$.

We need to select 4 letters from these 9 letters. We want to find the number of selections where the number of Cs is not the same as the number of Os. Let $n(C)$ be the number of Cs selected and $n(O)$ be the number of Os selected. We want to find the number of selections such that $n(C) \neq n(O)$.

It is easier to calculate the total number of ways to select 4 letters and subtract the number of ways where $n(C) = n(O)$.

First, calculate the total number of ways to select 4 letters from CROCODILE. We can categorize the selections based on the pattern of repeated letters:

1. Four distinct letters: The 7 distinct letters are C, O, R, D, I, L, E. We choose 4 distinct letters from these 7. The number of ways is $\binom{7}{4} = \frac{7 \times 6 \times 5 \times 4}{4 \times 3 \times 2 \times 1} = 35$.

2. Two alike of one kind and two distinct letters: The letters available with frequency 2 are C and O.

   • Choose C as the pair (CC): We need 2 more distinct letters from the remaining 6 distinct types {O, R, D, I, L, E}. Number of ways to choose these 2 letters is $\binom{6}{2} = 15$.
   • Choose O as the pair (OO): We need 2 more distinct letters from the remaining 6 distinct types {C, R, D, I, L, E}. Number of ways to choose these 2 letters is $\binom{6}{2} = 15$.

   Total ways for this case = $15 + 15 = 30$.

3. Two alike of one kind and two alike of another kind: The letters available with frequency 2 are C and O. We must choose both types to form the two pairs (CC and OO). There is only $\binom{2}{2} = 1$ way to choose the types, resulting in the selection CCOO. Number of ways = 1.

Total number of ways to select 4 letters = (Four distinct) + (Two alike, two distinct) + (Two pairs) = $35 + 30 + 1 = 66$.

Next, calculate the number of ways where $n(C) = n(O)$. The possible values for $n(C)$ and $n(O)$ are 0, 1, or 2 (since there are only two Cs and two Os available). The condition $n(C) = n(O)$ means the possible pairs $(n(C), n(O))$ are (0,0), (1,1), or (2,2). Let $n(X)$ be the number of letters selected from the set of single letters $S = \{R, D, I, L, E\}$. The total number of letters selected is 4, so $n(C) + n(O) + n(X) = 4$. The maximum number of letters we can select from $S$ is 5.

Case 1: $n(C) = 0$ and $n(O) = 0$. We select 0 Cs and 0 Os. We need to select 4 letters from the 5 distinct letters in $S$. Number of ways = $\binom{5}{4} = 5$.

Case 2: $n(C) = 1$ and $n(O) = 1$. We select 1 C (from the two available Cs) and 1 O (from the two available Os). We need to select $n(X) = 4 - n(C) - n(O) = 4 - 1 - 1 = 2$ letters from the 5 distinct letters in $S$. Number of ways to select 1 C = 1 (we simply choose one C). Number of ways to select 1 O = 1 (we simply choose one O). Number of ways to select 2 letters from $S = \binom{5}{2} = \frac{5 \times 4}{2} = 10$. Total ways for this case = $1 \times 1 \times 10 = 10$.

Case 3: $n(C) = 2$ and $n(O) = 2$. We select 2 Cs (from the two available Cs) and 2 Os (from the two available Os). We need to select $n(X) = 4 - n(C) - n(O) = 4 - 2 - 2 = 0$ letters from $S$. Number of ways to select 2 Cs = 1 (we take both Cs). Number of ways to select 2 Os = 1 (we take both Os). Number of ways to select 0 letters from $S = \binom{5}{0} = 1$. Total ways for this case = $1 \times 1 \times 1 = 1$.

The total number of selections where $n(C) = n(O)$ is the sum of ways in these three cases: $5 + 10 + 1 = 16$.

The number of selections in which the number of Cs is not the same as the number of Os is the total number of selections minus the number of selections where $n(C) = n(O)$. Number of selections with $n(C) \neq n(O) = 66 - 16 = 50$.

The final answer is 50.