The letters of the word EAMCET are permuted at random. The probability that the two E's will never be together, is

Options

A. `1/3`
B. `1/4`
C. `2/3`
D. `3/4`

Answer

To find the probability that the two E's in the word "EAMCET" will never be together, we can follow these steps:

### Step 1: Calculate the total number of arrangements of the letters in "EAMCET".

The word "EAMCET" consists of 6 letters where the letter E is repeated twice. The formula for the total arrangements of letters when there are repetitions is given by:

\[ \text{Total arrangements} = \frac{n!}{p_1! \times p_2! \times \ldots} \]

where \( n \) is the total number of letters and \( p_1, p_2, \ldots \) are the frequencies of the repeated letters.

For "EAMCET":
- Total letters, \( n = 6 \)
- The letter E appears 2 times.

Thus, the total arrangements are:

\[ \text{Total arrangements} = \frac{6!}{2!} = \frac{720}{2} = 360 \]

### Step 2: Calculate the number of arrangements where the two E's are together.

To find the arrangements where the two E's are together, we can treat the two E's as a single unit. Therefore, we can think of the arrangement as consisting of the units: (EE), A, M, C, T.

This gives us 5 units to arrange:

\[ \text{Units} = (EE), A, M, C, T \]

The number of arrangements of these 5 units is:

\[ \text{Arrangements with E's together} = 5! = 120 \]

### Step 3: Calculate the number of arrangements where the two E's are not together.

To find the number of arrangements where the two E's are not together, we subtract the number of arrangements where the E's are together from the total arrangements:

\[ \text{Arrangements with E's not together} = \text{Total arrangements} - \text{Arrangements with E's together} \]

\[ \text{Arrangements with E's not together} = 360 - 120 = 240 \]

### Step 4: Calculate the probability that the two E's will never be together.

The probability that the two E's are not together is given by the ratio of the number of favorable outcomes to the total outcomes:

\[ \text{Probability} = \frac{\text{Arrangements with E's not together}}{\text{Total arrangements}} = \frac{240}{360} \]

Simplifying this fraction:

\[ \text{Probability} = \frac{240 \div 120}{360 \div 120} = \frac{2}{3} \]

### Final Answer:
The probability that the two E's will never be together is \( \frac{2}{3} \).

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