To solve the problem of how many different words can be formed from the letters of the word "MALKENKOV" such that no two vowels are together, we can follow these steps:
### Step 1: Identify the letters and their types
The word "MALKENKOV" consists of the following letters:
- Consonants: M, L, K, N, K, V (total of 6 consonants)
- Vowels: A, E, O (total of 3 vowels)
### Step 2: Arrange the consonants
First, we need to arrange the consonants. The consonants are M, L, K, N, K, V. Note that 'K' appears twice.
The formula for arranging n items where there are repetitions is given by:
\[
\text{Number of arrangements} = \frac{n!}{p_1! \times p_2! \times \ldots}
\]
where \( p_1, p_2, \ldots \) are the frequencies of the repeated items.
In our case:
- Total consonants = 6 (M, L, K, N, K, V)
- Repeated consonants = K (2 times)
Thus, the number of arrangements of the consonants is:
\[
\text{Arrangements of consonants} = \frac{6!}{2!} = \frac{720}{2} = 360
\]
### Step 3: Determine the positions for the vowels
After arranging the consonants, we need to find the positions where we can place the vowels. When the consonants are arranged, they create gaps where vowels can be placed.
For example, if we have arranged the consonants as C1 C2 C3 C4 C5 C6, the gaps for placing vowels are:
- Before C1
- Between C1 and C2
- Between C2 and C3
- Between C3 and C4
- Between C4 and C5
- Between C5 and C6
- After C6
This gives us a total of 7 gaps.
### Step 4: Choose positions for the vowels
We need to choose 3 gaps from these 7 available gaps to place the vowels. The number of ways to choose 3 gaps from 7 is given by the combination formula:
\[
\binom{n}{r} = \frac{n!}{r!(n-r)!}
\]
where \( n \) is the total number of items to choose from, and \( r \) is the number of items to choose.
Thus, the number of ways to choose 3 gaps from 7 is:
\[
\binom{7}{3} = \frac{7!}{3!(7-3)!} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35
\]
### Step 5: Arrange the vowels
Now, we need to arrange the 3 vowels A, E, O. The number of arrangements of the vowels is given by:
\[
3! = 6
\]
### Step 6: Calculate the total arrangements
Finally, we can calculate the total number of arrangements by multiplying the arrangements of consonants, the ways to choose gaps, and the arrangements of vowels:
\[
\text{Total arrangements} = (\text{Arrangements of consonants}) \times (\text{Ways to choose gaps}) \times (\text{Arrangements of vowels})
\]
\[
\text{Total arrangements} = 360 \times 35 \times 6
\]
Calculating this gives:
\[
360 \times 35 = 12600
\]
\[
12600 \times 6 = 75600
\]
Thus, the total number of different words that can be formed from the letters of "MALKENKOV" such that no two vowels are together is **75,600**.
