Understanding Permutation and Combination of Alike Objects

Permutation and combination of alike objects addresses enumeration procedures where certain objects are indistinguishable from each other, thus modifying the standard formulas for arrangements and selections in finite sets.

Enumeration of Arrangements with Repeated Objects

Consider $n$ objects, out of which $n_1$ are of one kind, $n_2$ of another, ..., $n_k$ of a $k$th kind, and the rest are all distinct. The number of distinct permutations is given by

$\displaystyle \frac{n!}{n_1! \, n_2! \cdots n_k!}$

This formula is central for arrangements involving repeated letters or objects, such as in words with repeated alphabets.

Application to the Arrangement of Words with Repeated Letters

If a word contains letters, with some occurring more than once, the total permutations account for indistinguishability. For example, the word "BALLOON" has 7 letters with "L" repeated 2 times and "O" repeated 2 times. The total distinct arrangements are:

$\displaystyle \frac{7!}{2! \, 2!} = 1260$

Counting Selections of Alike Objects: Combinations with Repetition

Selections where each type of object may be chosen multiple times correspond to the "stars and bars" method. The number of ways to select $r$ objects from $k$ types, with unlimited supply, is:

$\displaystyle {k + r - 1 \choose r}$

This formula also describes the number of non-negative integer solutions to $x_1 + x_2 + \cdots + x_k = r$.

For syllabus-aligned revision material, refer to Permutations And Combinations Revision Notes.

Distinct Arrangement and Linear Permutations: Exam Patterns

In competitive examinations, calculation of linear arrangements with repeated elements is frequent. When all objects are distinct, total arrangements are $n!$; if repetitions occur, the denominator reduces the count due to indistinguishability.

For conceptual breadth, see Permutations And Combinations Overview.

Illustrative Cases: Worked Examples with Alike Objects

Example. In how many ways can the letters of "MISSISSIPPI" be arranged?

Total letters: $11$ ($M=1$, $I=4$, $S=4$, $P=2$)

Substitute into the formula:

$\dfrac{11!}{1! \times 4! \times 4! \times 2!} = \dfrac{39916800}{1 \times 24 \times 24 \times 2} = \dfrac{39916800}{1152} = 34650$

Example. In how many ways can 5 identical red balls, 3 identical blue balls, and 2 identical green balls be arranged in a row?

Total objects: $5+3+2=10,$ types: red, blue, green

Number of arrangements:

$\dfrac{10!}{5! \, 3! \, 2!} = \dfrac{3628800}{120 \times 6 \times 2} = \dfrac{3628800}{1440} = 2520$

Example. In how many ways can 3 alike mathematics books, 2 alike physics books, and 1 chemistry book be arranged on a shelf?

Number of books: $3+2+1=6$, indistinguishable books within the same subject

Arrangements:

$\dfrac{6!}{3! \, 2! \, 1!} = \dfrac{720}{6 \times 2 \times 1} = \dfrac{720}{12} = 60$

Example. Determine the number of ways to distribute 7 identical balls into 3 distinct boxes so that no box is empty.

This is equivalent to the number of positive integer solutions to $x_1 + x_2 + x_3 = 7$.

Number of solutions:

${7-1 \choose 3-1} = {6 \choose 2} = 15$

For problem-solving guidance, consult Permutation And Combination Problems.

Evaluating Multinomial Coefficients for Indistinguishable Objects

General division into $k$ groups of sizes $n_1, n_2, \ldots, n_k$ (with $n_1 + \cdots + n_k = n$) employs the multinomial coefficient:

$\displaystyle \frac{n!}{n_1! n_2! \ldots n_k!}$

This identity is applied whenever objects of various types, some alike, are to be arranged in order or assigned to groups.

Contrast with Selection of Distinct and Alike Objects

When only distinct objects are considered, the number of permutations and combinations utilizes $n!$ and $^nC_r$ directly. Inclusion of alike objects reduces arrangement count by accounting for indistinguishability.

Further detail is available at Important Questions On Permutations And Combinations.

Distinctions: Permutations vs. Combinations of Alike Objects

A deeper discussion on JEE-related function types can be found at Types Of Functions.

Common Student Errors in Arrangements with Alike Objects

• Ignoring divisor factorial for repeated objects
• Applying $n!$ directly when repetition exists
• Confusing selection with arrangement situations
• Neglecting the "at least one" condition in box problems
• Miscounting the total number of objects

For applied exercises in differentiation and combinations, also refer to Examples Of Derivatives.