 # Give full and correct answer in how many ways can a set of two positive integers less than 100 be chosen? 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.

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: $(4 – 1)!=3!=6$ Moreover, 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 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: $=24 – 12=12\space Ways$