This problem aims to familiarize us with **permutations **and **decision trees.** The concepts required to solve this problem are related to **algorithms** and **data structures** which include **computation, permutation, combination,** and **decision trees.**

In **data structures, permutation** correlates to the action of **organizing** all the components of a set into an **arrangement** or order. We can say that, if the set is already **ordered,** then the **rearranging** of its elements is called the process of **permitting.** A **permutation** is the selection of $r$ items from a set of $n$ items without a **substitute** and in order. Its **formula** is:

\[P^{n}_r = \dfrac{(n!)}{(n-r)!}\]

Whereas the **combination** is a method of choosing **entities** from a group, in which the arrangement of choice is not **important.** In shorter **combinations,** it is likely to estimate the number of **combinations.** A **combination** is the selection of $r$ items from a set of $n$ items without a substitute irrespective of the **arrangement:**

\[C^{n}_r =\dfrac{(P^{n}_r)}{(r!)}=\dfrac{(n!)}{r!(n-r)!}\]

## Expert Answer

Let’s consider that we have a **collection** of $n$ items. This implies that there are $n!$ **permutations** in which the **collection** can be organized.

Now a **decision tree** includes a **main** node, some **branches,** and **leaf** nodes. Every inner **node** represents a test, every **branch** represents the result of a test, and every **leaf** node carries a class label. We also know that a complete **decision tree** has $n!$ leaves but they are not **required** to be on the same **level.**

The **shortest possible answer** to the problem is $n − 1$. To briefly look at this, assume that we **carry** a **root-leaf** path let’s say $p_{r \longrightarrow l}$ with $k$ **comparisons,** we cannot be certain that the **permutation** $\pi (l)$ at the leaf $l$ is justified the correct **one.**

To **prove** this, consider a **tree** of $n$ nodes, where every **node** $i$ denotes $A[i]$. **Construct** an edge from $i$ to $j$ if we compare $A[i]$ with $A[j]$ on the track from the main **node** to $l$. Remark that for $k < n − 1$, this **tree** on ${1, . . . , n}$ will not be **combined.** Therefore, we have **two elements** $C_1$ and $C_2$ and we assume that nothing is known about the **comparative order** of **collection** items indexed by $C_1$ against items indexed by $C_2$.

Hence, there cannot exist a single **permutation** $\pi$ that arranges all **intakes** passing these $k$ tests – so $\pi (l)$ is inappropriate for some **collections** which guide to leaf $l$.

## Numerical Result

The **shortest** likely **depth** of a leaf in a **decision tree** for a **comparison** sort comes out to be $n − 1$.

## Example

Find the **number** of **ways** to arrange $6$ **children** in a line, if two individual children are constantly together.

According to the **statement,** $2$ students must be **together,** thus considering them as $1$.

Hence, the **outstanding** $5$ gives the **configuration** in $5!$ ways, i.e. $120$.

Furthermore, the $2$ children can be **organized** in $2!$ distinct ways.

Therefore, the **total** number of **arrangements** will be:

\[5!\times 2! = 120\times 2 = 240\space ways\]