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Question: Let \(A\) be the set of all positive prime numbers less than 30. Then find the number of different r...

Let AA be the set of all positive prime numbers less than 30. Then find the number of different rational numbers whose numerator and denominator belongs toAA.

Explanation

Solution

This question can be solved using the concept of combinations. Combinations in mathematics can be simply explained as a mathematical technique that determines the number of arrangements possible for collection of items where the order of the selection does not matter. So in combinations we can select an item in any order. So by using this concept we can solve the above given question.

Complete step-by-step solution:
Given
A:  Set of all positive prime numbers less than 30...............(i)A:\;{\text{Set of all positive prime numbers less than 30}}...............\left( i \right)
Now we have to find the number of different rational numbers whose numerator and denominator belongs toAA.
So first let us list all positive prime numbers less than 30. It can be written as:
A=2,3,5,7,11,13,17,19,23,29A = 2,3,5,7,11,13,17,19,23,29
So in total there would be 10 such numbers inAA.
Now we know that the basic definition of a rational number is:
A rational number is a number that can be in the form pq\dfrac{p}{q} where pp and qqare integers and qq is not equal to zero.
So using the theory of combinations we can write:
Number of waysppcan be selected fromAA: 10C1=10!1!(101)!=10......................(ii)^{10}{C_1} = \dfrac{{10!}}{{1!\left( {10 - 1} \right)!}} = 10......................\left( {ii} \right)
Number of waysqqcan be selected fromAA:(101)C1=9C1=9!1!(91)!=9...................(iii)^{\left( {10 - 1} \right)}{C_1}{ = ^9}{C_1} = \dfrac{{9!}}{{1!\left( {9 - 1} \right)!}} = 9...................\left( {iii} \right)
So from (ii) and (iii) the number of different rational numbers whose numerator and denominator belongs toAAcan be written as:
10×9=90.................(iv)10 \times 9 = 90.................\left( {iv} \right)

Therefore our final answer would be: 9090

Note: General formula to find combination in the formnCr^n{C_r}is:
nCr=n!r!(nr)!^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}
While doing a question involving combinations one should be careful about the calculation processes since it can be tricky as well as lengthy.
Also the above given method is recommended to solve similar questions involving combinations.