How many words can be formed with the letters of the word 'DELHI' if E and H never occur together?

Options

A. i) 6
B. ii) 72
C. iii) 24
D. iv) 60

Answer

To solve the problem of how many words can be formed with the letters of the word 'DELHI' if E and H never occur together, we can follow these steps:

### Step 1: Calculate the total arrangements of the letters in 'DELHI'.
The word 'DELHI' consists of 5 distinct letters: D, E, L, H, I. The total number of arrangements (permutations) of these 5 letters is given by:

\[ 5! = 120 \]

### Step 2: Calculate the arrangements where E and H are together.
To find the arrangements where E and H are together, we can treat E and H as a single unit or block. This means we can consider the block (EH) as one letter. Therefore, we now have the following letters to arrange: (EH), D, L, I. This gives us a total of 4 units to arrange.

The number of arrangements of these 4 units is:

\[ 4! = 24 \]

Since E and H can be arranged among themselves in 2 ways (EH or HE), we multiply the arrangements of the 4 units by the arrangements of E and H:

\[ 4! \times 2! = 24 \times 2 = 48 \]

### Step 3: Calculate the arrangements where E and H are NOT together.
To find the arrangements where E and H are not together, we subtract the number of arrangements where E and H are together from the total arrangements:

\[ \text{Arrangements where E and H are NOT together} = 5! - (4! \times 2!) \]

Substituting the values we calculated:

\[ = 120 - 48 = 72 \]

### Final Answer:
Therefore, the number of words that can be formed with the letters of the word 'DELHI' such that E and H never occur together is **72**.

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