Permutations with Identical Objects & Multinomial Theorem

Concept Overview

This question tests the fundamental concept of permutations, specifically how to handle arrangements when some objects are identical. We'll start with the basic formula for permutations with one group of identical items and then generalize this using the multinomial theorem to solve problems involving multiple groups of identical items. This is crucial for understanding complex counting problems in combinatorics.

Step 1: Understand permutations with identical objects. When arranging $n$ objects where $k$ of them are identical, the total number of distinct permutations is given by: $\frac{n!}{k!}$ This formula arises because if all $n$ objects were distinct, there would be $n!$ permutations. However, since $k$ objects are identical, swapping them among themselves does not create a new distinct arrangement. There are $k!$ ways to arrange these $k$ identical objects, so we divide the total permutations by $k!$ to correct for overcounting.

Step 2: Extend to multiple groups of identical objects (Multinomial Theorem application). Consider arranging $n$ objects where there are $k_1$ identical objects of type 1, $k_2$ identical objects of type 2, ..., $k_m$ identical objects of type $m$, such that $k_1 + k_2 + \dots + k_m = n$. The number of distinct permutations is given by the multinomial coefficient: $\frac{n!}{k_1! k_2! \dots k_m!}$ This is a direct extension of the previous case. For each group of identical objects, we divide by the factorial of the count of those identical objects to avoid overcounting. The multinomial theorem provides the coefficients for the expansion of $(x_1 + x_2 + \dots + x_m)^n$, and these coefficients are precisely the number of ways to partition $n$ distinct items into $m$ distinct groups of sizes $k_1, k_2, \dots, k_m$. In permutation problems, we are essentially distributing $n$ positions among these groups.

Step 3: Apply the formula to a sample problem. Let's find the number of distinct permutations of the letters in the word "MISSISSIPPI". Here, $n = 11$ (total number of letters). The counts of identical letters are:

• M: $k_1 = 1$
• I: $k_2 = 4$
• S: $k_3 = 4$
• P: $k_4 = 2$ We check that $k_1 + k_2 + k_3 + k_4 = 1 + 4 + 4 + 2 = 11 = n$.

Step 4: Calculate the number of permutations. Using the multinomial formula: $\frac{11!}{1! 4! 4! 2!}$ Calculate the factorials: $11! = 39,916,800$ $1! = 1$ $4! = 4 \times 3 \times 2 \times 1 = 24$ $2! = 2 \times 1 = 2$

Substitute these values into the formula: $\frac{39,916,800}{1 \times 24 \times 24 \times 2} = \frac{39,916,800}{1152}$ $= 34,650$ Therefore, there are 34,650 distinct permutations of the letters in "MISSISSIPPI".

Key Takeaways:

• When arranging $n$ objects with $k$ identical items, the number of distinct permutations is $n!/k!$.
• For $n$ objects with multiple groups of identical items ($k_1, k_2, \dots, k_m$), the number of distinct permutations is $n!/(k_1! k_2! \dots k_m!)$.
• This formula is a direct application of the multinomial coefficient, used in combinatorics for partitioning and distribution problems.
• Always ensure the sum of the counts of identical items equals the total number of items ($k_1 + k_2 + \dots + k_m = n$).

Answer: 34,650