The letters of the word NAVANAVALAVANYAM are arranged in a row at random. The probability that repeated letters of the same kind are together, is

Options

A. `(6(7!)(3!)^(2))/(16!)`
B. `((6!)^(2)(3!)^(2))/(16!)`
C. `((7!)^(2)(3!)^(2))/(16!)`
D. `(6!7!(3!)^(2))/(16!)`

Answer

To solve the problem of finding the probability that repeated letters of the same kind in the word "NAVANAVALAVANYAM" are together, we can follow these steps:

### Step 1: Count the total number of letters and their frequencies.
The word "NAVANAVALAVANYAM" consists of 16 letters. We can break down the letters as follows:
- N: 3 times
- A: 7 times
- V: 3 times
- L: 1 time
- M: 1 time
- Y: 1 time

### Step 2: Calculate the total arrangements of the letters.
The total number of arrangements of the letters can be calculated using the formula for permutations of multiset:

\[ \text{Total arrangements} = \frac{n!}{n_1! \cdot n_2! \cdot n_3! \cdots} \]

Where \( n \) is the total number of letters and \( n_1, n_2, n_3, \ldots \) are the frequencies of each letter.

So, we have:

\[ \text{Total arrangements} = \frac{16!}{3! \cdot 7! \cdot 3! \cdot 1! \cdot 1! \cdot 1!} \]

### Step 3: Calculate the arrangements with repeated letters together.
To find the arrangements where the repeated letters are together, we can treat each group of repeated letters as a single unit.

- For N (3 N's together), we consider it as one unit: NNN
- For A (7 A's together), we consider it as one unit: AAAAAAA
- For V (3 V's together), we consider it as one unit: VVV

Now we have:
- 1 unit of NNN
- 1 unit of AAAAAAA
- 1 unit of VVV
- 1 unit of L
- 1 unit of M
- 1 unit of Y

This gives us a total of 6 units to arrange.

The number of arrangements of these 6 units is:

\[ \text{Arrangements with letters together} = 6! \]

### Step 4: Calculate the probability.
The probability that the repeated letters are together is given by the ratio of the number of favorable arrangements to the total arrangements:

\[ P(\text{repeated letters together}) = \frac{\text{Arrangements with letters together}}{\text{Total arrangements}} = \frac{6!}{\frac{16!}{3! \cdot 7! \cdot 3! \cdot 1! \cdot 1! \cdot 1!}} \]

This simplifies to:

\[ P(\text{repeated letters together}) = \frac{6! \cdot 3! \cdot 7! \cdot 3! \cdot 1! \cdot 1! \cdot 1!}{16!} \]

### Step 5: Final Calculation.
Now we can compute the values:

1. \( 6! = 720 \)
2. \( 3! = 6 \)
3. \( 7! = 5040 \)
4. \( 16! = 20922789888000 \)

Thus,

\[ P(\text{repeated letters together}) = \frac{720 \cdot 6 \cdot 5040 \cdot 6}{20922789888000} \]

Calculating the numerator:

\[ 720 \cdot 6 = 4320 \]
\[ 4320 \cdot 5040 = 21772800 \]
\[ 21772800 \cdot 6 = 130636800 \]

So,

\[ P(\text{repeated letters together}) = \frac{130636800}{20922789888000} \]

This fraction can be simplified further to find the final probability.