Question

Number of positive integral solutions of \[xyz = 30\] is
(a) 9
(b) 27
(c) 81
(d) 243

Answer 1

Hint: We will begin by writing all the possible forms when we can 30 as a product of 3 numbers. Then, we will find the number of ways in which the variables, \[x,y,z\] can take the respective values using the rule if there are $n$ distinct objects and $n$places, then there are $n!$ ways. At last, add all the number of ways to get the required answer.

Complete step-by-step answer:
First of all, we will write all the possible ways we can write the product three numbers as 30.
$30 = 1 \times 5 \times 6$,
$30 = 1 \times 2 \times 15$
$30 = 1 \times 3 \times 10$
$30 = 2 \times 3 \times 5$
And $30 = 1 \times 1 \times 30$
Any of the variables, \[x,y,z\] can have any value from the above values.
And if there are $n$ distinct objects and $n$ places, then there are $n!$ ways
The number of ways in which \[x,y,z\] can take values 1,5,6 is 3!
Similarly, the number of ways in which \[x,y,z\] can take values 1,2,15 is 3!
The number of ways in which \[x,y,z\] can take values 1,3,10 is 3!
And the number of ways in which \[x,y,z\] can take values 2,3,5 is 3!
It is known that if there are $n$ objects, and object 1 is repeated ${n_1}$ times, then the number of arrangements are $\dfrac{{n!}}{{{n_1}!}}$
The number of ways in which \[x,y,z\] can take values 1,1,30 is \[\dfrac{{3!}}{{2!}}\]
We will add all the possible cases to get the total number of ways.
$3! + 3! + 3! + 3! + \dfrac{{3!}}{{2!}}$
Also, $n! = n\left( {n - 1} \right)\left( {n - 2} \right).....3.2.1$
Then,
$
  3.2.1 + 3.2.1 + 3.2.1 + 3.2.1 + \dfrac{{3.2!}}{{2!}} \\
   \Rightarrow 6 + 6 + 6 + 6 + 3 = 27 \\
$
Hence, option (b) is correct.

Note: Many students write the number of ways in which \[x,y,z\] can take values 1,1,30 is 3!, which is incorrect. Since, all the numbers are not distinct, we will divide it by the number of times an object is repeated.