Permutation under Condition

Permutations under condition refer to the arrangements of objects in a specific order, where certain conditions or restrictions are applied. These conditions can vary widely, from objects needing to be adjacent to each other, to certain objects not being allowed to be in specific positions.

Understanding Permutations

Before diving into permutations under conditions, let's first understand what a permutation is. A permutation is an arrangement of all or part of a set of objects, with regard to the order of the arrangement. For example, the permutations of the set {A, B, C} are ABC, ACB, BAC, BCA, CAB, and CBA.

The number of permutations of n distinct objects taken r at a time is given by the formula:

[ P(n, r) = \frac{n!}{(n-r)!} ]

where n! (n factorial) is the product of all positive integers up to n.

Permutations with Conditions

When conditions are applied, the calculation of permutations becomes more complex. The conditions can be of various types, such as 'certain objects must be together', 'certain objects must not be together', 'objects are not distinct', or 'positions are restricted for certain objects'.

Table of Differences and Important Points

Condition Type	Description	Formula (if applicable)	Example
Objects Together	Certain objects must be adjacent in the arrangement.	Use block method.	Permute "AB" as a single block with other elements.
Objects Apart	Certain objects must not be adjacent.	Use subtraction method.	Total permutations - permutations with objects together.
Non-Distinct Objects	There are repeating objects in the set.	[ \frac{n!}{n_1! \cdot n_2! \cdot \ldots \cdot n_k!} ]	Permute "AAB".
Restricted Positions	Certain objects can only be in specific positions.	Case-by-case basis.	"A" must be at the first position.

Examples to Explain Important Points

Example 1: Objects Together

Suppose we want to find the number of ways to arrange the letters in the word "BOND" such that "B" and "O" are always together.

We can treat "BO" as a single block, and thus we have three entities to arrange: [BO], N, D.

The number of permutations of these three entities is:

[ P(3, 3) = 3! = 6 ]

However, within the block [BO], "B" and "O" can be arranged in 2! ways. Therefore, the total permutations are:

[ 3! \times 2! = 6 \times 2 = 12 ]

Example 2: Objects Apart

Let's find the number of ways to arrange the letters in the word "BEAD" such that "B" and "E" are never together.

First, find the total number of permutations without any restrictions:

[ P(4, 4) = 4! = 24 ]

Next, find the number of permutations where "B" and "E" are together (as in Example 1):

[ 3! \times 2! = 6 \times 2 = 12 ]

Now, subtract the permutations with "B" and "E" together from the total permutations:

[ 24 - 12 = 12 ]

So, there are 12 arrangements where "B" and "E" are not together.

Example 3: Non-Distinct Objects

Find the number of ways to arrange the letters in the word "MISSISSIPPI".

Here, we have repeating letters: M(1), I(4), S(4), P(2).

Using the formula for non-distinct objects:

[ \frac{n!}{n_1! \cdot n_2! \cdot \ldots \cdot n_k!} = \frac{11!}{1! \cdot 4! \cdot 4! \cdot 2!} ]

Calculating this gives us the number of distinct permutations.

Example 4: Restricted Positions

How many ways can we arrange the letters in the word "CABLE" if "C" must be at the beginning?

Since "C" is fixed at the first position, we only need to arrange "ABLE".

[ P(4, 4) = 4! = 24 ]

So, there are 24 arrangements with "C" at the beginning.

Conclusion

Permutations under conditions require careful consideration of the restrictions imposed. By understanding the different types of conditions and applying the appropriate methods and formulas, one can calculate the number of permissible arrangements. Practice with various conditions is essential to master this topic, which is crucial for exams involving permutations and combinations.