To solve the problem of arranging the letters of the word "COCONUT" such that at least one 'C' comes at an odd position, we can follow these steps:
### Step 1: Calculate the total arrangements of the letters in "COCONUT" without any restrictions.
The word "COCONUT" consists of 7 letters where:
- C appears 2 times,
- O appears 2 times,
- N appears 1 time,
- U appears 2 times,
- T appears 1 time.
The formula for the 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.
Thus, we have:
\[
\text{Total arrangements} = \frac{7!}{2! \times 2! \times 2! \times 1!}
\]
Calculating this gives:
\[
= \frac{5040}{2 \times 2 \times 2} = \frac{5040}{8} = 630
\]
### 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). The even positions are 2, 4, and 6 (3 positions).
If we want to ensure that no 'C' is at an odd position, we can only place 'C's in the even positions (2, 4, 6). We need to choose 2 out of these 3 even positions for the 'C's.
The number of ways to choose 2 positions from 3 is given by:
\[
\binom{3}{2} = 3
\]
Now, we have 5 remaining letters (O, O, N, U, U) to arrange in the remaining 5 positions (1 odd and 2 even). The arrangements of these letters is given by:
\[
\text{Arrangements of remaining letters} = \frac{5!}{2! \times 2! \times 1!}
\]
Calculating this gives:
\[
= \frac{120}{2 \times 2} = \frac{120}{4} = 30
\]
### Step 3: Calculate the total arrangements where no 'C' is at an odd position.
Now, we multiply the number of ways to choose positions for 'C's with the arrangements of the remaining letters:
\[
\text{Total arrangements with no 'C' at odd positions} = 3 \times 30 = 90
\]
### Step 4: 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:
\[
\text{Arrangements with at least one 'C' at odd position} = \text{Total arrangements} - \text{Arrangements with no 'C' at odd positions}
\]
Substituting the values we calculated:
\[
= 630 - 90 = 540
\]
### 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 **540**.
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