The aim of this problem is to introduce us to **Combination **and **Permutation.** The concepts required to solve this problem are related to** discrete mathematics. ****Combination** and **permutation** are two distinct **methods** of grouping items of a set into subsets. If the **items** are of the subset are arranged in any order, then its **combination,** otherwise, if it is arranged in order, then it’s **permutation.**

In mathematics, a **combination** is the organization of its given elements irrespective of their **arrangement.**

\[C^{n}_{r}=\dfrac{n!}{r!(n-r)!} \space \text{where} 0 \leq r \leq n\]

Where, $C^{n}_{r}$**= Combinations,**

$n$ =** Total elements,**

$r$ **= Desired elements,**

Furthermore, a **permutation** is the engagement of its elements in a **defined order.** The arrangement of the elements is a key aspect of **permutation** also known as an **ordered combination.** It is given by:

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

Where $P^{n}_{r}$ is the total **number** of **permutations,**

$n$ =** Total elements,**

$r$ = **Desired elements.**

## Expert Answer

According to the problem **statement,** we are given a set of $2$ **positive integers,** and we have to find the **number** of **ways** in which the chosen **integers** will be less than $100$.

Since the **order** of the **numbers** in a set is not important, as a different **arrangement** of elements in a set remains the same **set,** thus we will be using the concept of **combination.**

Also in a set of $100$ **numbers,** there are $99$ elements that are **less** than the $100$ **value** mark, and of which we are to **select** only $2$ **numbers. Thus:**

\[n=\text{less than 100}=99\]

\[r=\text{selected}=2\]

**Substituting** the values in $C^{n}_{r}=\dfrac{n!}{r!(n-r)!}$,

\[C^{99}_{2}=\dfrac{99!}{2!(99-2)!}\]

The **factorial** of $99$ is $(99\times 98\times 97\times …. 3\times 2\times 1)$.

\[C^{99}_{2}=\dfrac{(99\times 98\times …. 2\times 1)}{(2\times 1)(97)!}\]

The **factorial** of $97$ is $(97\times 96\times …. 3\times 2\times 1)$.

\[C^{99}_{2}=\dfrac{(99\times 98\times …. 2\times 1)}{(2\times 1)(97\times 96\times …. 3\times 2\times 1)}\]

\[C^{99}_{2}=\dfrac{9.332\times 10^{155}}{(2)(9.623\times 10^{151})}\]

\[C^{99}_{2}=\dfrac{9702}{2}\]

\[C^{99}_{2}=4851\space Ways\]

## Numerical Result

The **number** of **ways** in which a set of **two positive integers** that are less than 100 will be **chosen** is $4851\space Ways$.

## Example

Calculate the **number** of **ways** in which **five** persons **A, B, C, D,** and** E** sit around a round table such that:

- No
**restriction**occurs. - Person
**A**and**D**should be**seated together.** - Person
**C**and**E**can**not**be**seated**together.

The **number** of **ways,** $5$ people can be seated around a round table **without restrictions** =$(n-1)!$.

Total **number** of **ways** = $(n-1)!$,

\[(5 – 1)!=4!\]

\[=4\times 3\times 2\times 1\]

\[24\space\text{possible ways.}\]

For the **second** part, if persons **A** and **D** are to be **seated** together, we have to consider them as **one unit.** So the **arrangement** is to be done in $4$ ways:

\[(4 â€“ 1)!=3!=6\]

Moreover,Â **persons** **A** and **D** can shuffle their **positions** in $2$ ways. So the **arrangements** can be made in $6 \times 2 = 12$ ways.

In the **second**Â **part,** the number of ways **two persons** can sit is $12$ and the **total number** of **ways** was $24$. So the total number of **ways** in which persons C and E must not **sit together** is:

\[=24 â€“ 12=12\space Ways\]