Number of ways in which the letters of the word DECISIONS be arranged so that letter N be somewhere to the right of the letter "D" is `(lfloor9)/(lambda)`. Find `lambda`.

Answer

To solve the problem of arranging the letters of the word "DECISIONS" such that the letter 'N' is somewhere to the right of the letter 'D', we can follow these steps:

### Step 1: Count the total letters
The word "DECISIONS" consists of 9 letters: D, E, C, I, S, I, O, N, S.

### Step 2: Identify the arrangement condition
We need to ensure that 'N' is to the right of 'D'.

### Step 3: Calculate total arrangements without restrictions
The total arrangements of the letters in "DECISIONS" can be calculated using the formula for permutations of multiset:
\[ \text{Total arrangements} = \frac{9!}{2! \times 2!} \]
Here, we divide by \(2!\) for the two 'I's and \(2!\) for the two 'S's.

### Step 4: Calculate the arrangements with 'N' to the right of 'D'
Since 'N' must be to the right of 'D', we can consider that in half of the arrangements, 'N' will be to the right of 'D' and in the other half, 'N' will be to the left of 'D'. Thus, the arrangements where 'N' is to the right of 'D' will be:
\[ \text{Arrangements with N right of D} = \frac{1}{2} \times \text{Total arrangements} \]

### Step 5: Substitute the total arrangements
Substituting the total arrangements into the equation:
\[ \text{Arrangements with N right of D} = \frac{1}{2} \times \frac{9!}{2! \times 2!} \]

### Step 6: Simplify the expression
Calculating the above expression:
\[ = \frac{9!}{2! \times 2! \times 2} \]
This can be simplified to:
\[ = \frac{9!}{4!} \]

### Step 7: Relate to the given form
According to the problem, the number of ways is given by \(\frac{9!}{\lambda}\). Thus, we can equate:
\[ \lambda = 4! \]

### Step 8: Calculate \(4!\)
Calculating \(4!\):
\[ 4! = 24 \]

### Conclusion
Thus, the value of \(\lambda\) is:
\[ \lambda = 24 \]