**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.

**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: \[=24 â€“ 12=12\space Ways\]

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