Question

Number of positive integral solution of $xyz = 30$ is
(a) 9
(b)27
(c)81
(d)243

Answer 1

Hint: First we need to find values of x, y and z such that their product is equal to 30. Then use permutation to calculate the number of possible arrangements of these numbers.

Complete step-by-step answer:
Given the problem, we need to find the number of positive integral solutions of $xyz = 30$.
It means that we need to find the permutation of the number of values of $x,y,z$such that the above equation is satisfied.
First, we need to find the factors of 30.
Factors of 30 are 1,2,3,5,6,10,15 and 30.
From the given factors, we need to find sets of 3 factors such that their product is 30.
The sets satisfying above conditions would be
$
  x = 30,y = 1,z = 1 \\
  x = 15,y = 2,z = 1 \\
  x = 10,y = 3,z = 1 \\
  x = 6,y = 5,z = 1 \\
  x = 5,y = 3,z = 2 \\
$
The values of $x,y,z$can be interchanged as the multiplication is commutative in real numbers.
We know that number of ways to arrange n objects at a time $ = n!$
Also, the number of ways to arrange n objects at a time with r objects repeated $ = \dfrac{{n!}}{{r!}}$
Using the above permutation formula in the obtained sets, we get to find the number of viable or ways.
$
  x = 30,y = 1,z = 1{\text{ , no}}{\text{. of ways = }}\dfrac{{3!}}{{2!}} = 3 \\
  x = 15,y = 2,z = 1{\text{ , no}}{\text{. of ways = 3!}} = 6 \\
  x = 10,y = 3,z = 1{\text{ , no}}{\text{. of ways = 3!}} = 6 \\
  x = 6,y = 5,z = 1{\text{ , no}}{\text{. of ways = 3!}} = 6 \\
  x = 5,y = 3,z = 2{\text{ , no}}{\text{. of ways = 3!}} = 6 \\
$

This gives a total no. of ways = 6+6+6+6+3=27.
Hence the total number of ways in which product of $xyz = 30$is 27.
Hence no. of positive integral solutions of $xyz = 30$is 27.
Therefore, option (B) 27 is the correct answer.

Note: While counting the possible combination of factors in the above problem, the factors could be chosen such that the factor on the left is always greater than or equal to the factor on its right, not less. This approach could help in saving time and rectifying the redundant cases. Formula used for arrangement of objects should be kept in mind while solving problems like above.