To solve the problem of arranging the letters of the word "COCONUT" such that at least one 'C' occupies an odd position, we can follow these steps:
### Step 1: Calculate the total arrangements of the letters in "COCONUT".
The word "COCONUT" consists of 7 letters where:
- C appears 2 times,
- O appears 2 times,
- N appears 1 time,
- U appears 1 time,
- T appears 1 time.
The formula for the number of arrangements of letters when there are identical letters is given by:
\[
\text{Total arrangements} = \frac{n!}{p_1! \times p_2! \times \ldots \times p_k!}
\]
Where \( n \) is the total number of letters, and \( p_1, p_2, \ldots, p_k \) are the frequencies of the identical letters.
For "COCONUT":
\[
\text{Total arrangements} = \frac{7!}{2! \times 2! \times 1! \times 1! \times 1!}
\]
Calculating this gives:
\[
= \frac{5040}{2 \times 2 \times 1 \times 1 \times 1} = \frac{5040}{4} = 1260
\]
### Step 2: Calculate the arrangements where no 'C' is at an odd position.
The odd positions in a 7-letter arrangement are 1, 3, 5, and 7 (4 positions). Since we want to ensure that no 'C' is in these positions, we can only place 'C's in the even positions (2, 4, 6).
We can choose 2 out of the 3 even positions (2, 4, 6) for the 'C's. The number of ways to choose 2 positions from 3 is given by:
\[
\binom{3}{2} = 3
\]
After placing the 'C's in the chosen positions, we have 5 remaining letters (O, O, N, U, T) to arrange in the remaining 5 positions.
The arrangements of these 5 letters (considering O is identical) is given by:
\[
\text{Arrangements of O, O, N, U, T} = \frac{5!}{2!} = \frac{120}{2} = 60
\]
So, the total arrangements where no 'C' is at an odd position is:
\[
\text{Total arrangements with no C at odd positions} = 3 \times 60 = 180
\]
### Step 3: Calculate the arrangements where at least one 'C' is at an odd position.
To find the arrangements where at least one 'C' is at an odd position, we subtract the arrangements where no 'C' is at an odd position from the total arrangements:
\[
\text{Arrangements with at least one C at odd positions} = \text{Total arrangements} - \text{Arrangements with no C at odd positions}
\]
Substituting the values we calculated:
\[
= 1260 - 180 = 1080
\]
### Final Answer:
The number of ways in which all the letters of the word "COCONUT" can be arranged such that at least one 'C' comes at an odd place is **1080**.
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