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:
\[(4 – 1)!=3!=6\]
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:
\[=24 – 12=12\space Ways\]
