Answer
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Hint: Rational numbers are those numbers which can be expressed in the form of $\dfrac{p}{q}$ where q≠0. In the given integer, if the factors of denominator of the given rational number $\dfrac{p}{q}$ is in the form of ${2^m}{5^n}$, where $m$ and $n$ are non-negative integer, then the decimal expression of the rational number is terminating otherwise they will not terminating and they repeat continuously.
Complete step by step solution:
Given,
Total digits given=6
The given digits can occupy the numerators and denominators places in ways, $^6{P_2}$
We simplify the $^6{P_2}$
\[ = \dfrac{{6!}}{{(6 - 2)!}}\]
\[ = \dfrac{{6 \times 5 \times 4!}}{{4!}}\]
\[ = 30\]
There are 30 possible ways.
$\left\{ {\dfrac{1}{1},\dfrac{1}{2},\dfrac{1}{3},\dfrac{1}{4},\dfrac{1}{5},\dfrac{1}{6},\dfrac{2}{1},\dfrac{2}{2},\dfrac{2}{3},\dfrac{2}{4},\dfrac{2}{5},\dfrac{2}{6},\dfrac{3}{1},\dfrac{3}{2},\dfrac{3}{3},\dfrac{3}{4},\dfrac{3}{5},\dfrac{3}{6},\dfrac{4}{1},\dfrac{4}{2},\dfrac{4}{3},\dfrac{4}{4},\dfrac{4}{5},\dfrac{4}{6},\dfrac{5}{1},\dfrac{5}{2},\dfrac{5}{3},\dfrac{5}{4},\dfrac{5}{5},\dfrac{5}{6},\dfrac{6}{1},\dfrac{6}{2},\dfrac{6}{3},\dfrac{6}{4},\dfrac{6}{5},\dfrac{6}{6}} \right\}$
In which these number represent same,
$\left\{ {\dfrac{1}{2},\dfrac{2}{4},\dfrac{3}{6}} \right\}$,$\left\{ {\dfrac{2}{1},\dfrac{4}{2},\dfrac{6}{3}} \right\}$,$\left\{ {\dfrac{2}{3},\dfrac{4}{6}} \right\}$,$\left\{ {\dfrac{3}{2},\dfrac{6}{4}} \right\}$,$\left\{ {\dfrac{1}{3},\dfrac{2}{6}} \right\}$ and $\left\{ {\dfrac{3}{1},\dfrac{6}{2}} \right\}$
In these case we consider only one case
$ = no.\,of\,rational\,number$
$ = 30 - (2 + 2 + 1 + 1 + 1 + 1) + 1$
1 is a rational number.
$ = 31 - 8$
$ = 23$
Therefore, The number of rational numbers $\dfrac{p}{q}$where $p,q \in 1,2,3,4,5,6$ is 23. So, the correct option is (A).
Note:
The number system or numeral system is the system of naming. There are various types of number systems. Natural numbers are those used for counting. Whole numbers are those in zero including with natural numbers. Integers are those which include positive and negative numbers.
Complete step by step solution:
Given,
Total digits given=6
The given digits can occupy the numerators and denominators places in ways, $^6{P_2}$
We simplify the $^6{P_2}$
\[ = \dfrac{{6!}}{{(6 - 2)!}}\]
\[ = \dfrac{{6 \times 5 \times 4!}}{{4!}}\]
\[ = 30\]
There are 30 possible ways.
$\left\{ {\dfrac{1}{1},\dfrac{1}{2},\dfrac{1}{3},\dfrac{1}{4},\dfrac{1}{5},\dfrac{1}{6},\dfrac{2}{1},\dfrac{2}{2},\dfrac{2}{3},\dfrac{2}{4},\dfrac{2}{5},\dfrac{2}{6},\dfrac{3}{1},\dfrac{3}{2},\dfrac{3}{3},\dfrac{3}{4},\dfrac{3}{5},\dfrac{3}{6},\dfrac{4}{1},\dfrac{4}{2},\dfrac{4}{3},\dfrac{4}{4},\dfrac{4}{5},\dfrac{4}{6},\dfrac{5}{1},\dfrac{5}{2},\dfrac{5}{3},\dfrac{5}{4},\dfrac{5}{5},\dfrac{5}{6},\dfrac{6}{1},\dfrac{6}{2},\dfrac{6}{3},\dfrac{6}{4},\dfrac{6}{5},\dfrac{6}{6}} \right\}$
In which these number represent same,
$\left\{ {\dfrac{1}{2},\dfrac{2}{4},\dfrac{3}{6}} \right\}$,$\left\{ {\dfrac{2}{1},\dfrac{4}{2},\dfrac{6}{3}} \right\}$,$\left\{ {\dfrac{2}{3},\dfrac{4}{6}} \right\}$,$\left\{ {\dfrac{3}{2},\dfrac{6}{4}} \right\}$,$\left\{ {\dfrac{1}{3},\dfrac{2}{6}} \right\}$ and $\left\{ {\dfrac{3}{1},\dfrac{6}{2}} \right\}$
In these case we consider only one case
$ = no.\,of\,rational\,number$
$ = 30 - (2 + 2 + 1 + 1 + 1 + 1) + 1$
1 is a rational number.
$ = 31 - 8$
$ = 23$
Therefore, The number of rational numbers $\dfrac{p}{q}$where $p,q \in 1,2,3,4,5,6$ is 23. So, the correct option is (A).
Note:
The number system or numeral system is the system of naming. There are various types of number systems. Natural numbers are those used for counting. Whole numbers are those in zero including with natural numbers. Integers are those which include positive and negative numbers.
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