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Q11(iii):
In how many ways can the letters of the word PERMUTATIONS be arranged if the (iii) there are always 4 letters between P and S?

Solution :

Step 1: Analyis of the Given Word

The word "PERMUTATIONS" consists of $12$ letters. Let us identify the frequency of each letter:

  • P: 1, E: 1, R: 1, M: 1, U: 1, T: 2, A: 1, I: 1, O: 1, N: 1, S: 1.
  • Total letters ($n$) = 12.
  • Repeating letters: 'T' appears 2 times.

Step 2: Defining the Constraint

We are required to arrange the letters such that there are exactly 4 letters between 'P' and 'S'.

Let the positions be represented by indices $1, 2, 3, \dots, 12$. If P is at position $i$, then S must be at position $i+5$ (to allow for 4 letters between them) or vice versa. The possible pairs of positions $(P, S)$ are:

$(1, 6), (2, 7), (3, 8), (4, 9), (5, 10), (6, 11), (7, 12)$.

There are $7$ such pairs. Since P and S can be swapped, there are $7 \times 2 = 14$ ways to place P and S.

Step 3: Calculating Permutations of Remaining Letters

Once P and S are fixed in a chosen pair of positions, there are $12 - 2 = 10$ remaining positions to be filled by the remaining $10$ letters.

The remaining letters are $\{E, R, M, U, T, T, A, I, O, N\}$.

The number of ways to arrange these 10 letters, where 'T' repeats 2 times, is given by the formula for permutations of multisets [Per Fundamental Principle of Counting]:

$$\text{Ways} = \frac{10!}{2!}$$

Calculating the value:

$$\frac{3,628,800}{2} = 1,814,400$$

Step 4: Final Computation

The total number of arrangements is the product of the number of ways to place P and S and the number of ways to arrange the remaining letters.

$$\text{Total Arrangements} = (\text{Ways to place P and S}) \times (\text{Ways to arrange remaining 10 letters})$$

$$\text{Total Arrangements} = 14 \times 1,814,400$$

$$\text{Total Arrangements} = 25,401,600$$

Visual Representation of Position Slots

P 4 Letters S Constraint: Fixed distance of 5 units (4 letters between)

Final Solution: The total number of ways to arrange the letters such that there are always 4 letters between P and S is 25,401,600.


More Questions from Class 11 Mathematics Permutations and Combinations EXERCISE 6.3


CBSE Solutions for Class 11 Mathematics Permutations and Combinations


Chapters in CBSE - Class 11 Mathematics


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