The total number of positive integral solutions (x, y, z) such that xyz = 24 is :

Options

A. 36
B. 24
C. 45
D. 30

Answer

To find the total number of positive integral solutions (x, y, z) such that \(xyz = 24\), we can follow these steps:

### Step 1: Factorization of 24
First, we need to factor the number 24 into its prime factors:
\[ 24 = 2^3 \times 3^1 \]
This means we have three 2's and one 3 to distribute among the variables \(x\), \(y\), and \(z\).

### Step 2: Distributing the Factors
We need to distribute the prime factors \(2^3\) and \(3^1\) among \(x\), \(y\), and \(z\).

#### Distribution of \(2^3\):
Let \(x = 2^{a_1}\), \(y = 2^{a_2}\), \(z = 2^{a_3}\) where \(a_1 + a_2 + a_3 = 3\) and \(a_1, a_2, a_3 \geq 0\).

The number of non-negative integer solutions to the equation \(a_1 + a_2 + a_3 = 3\) can be found using the "stars and bars" theorem:
\[ \text{Number of solutions} = \binom{3 + 3 - 1}{3 - 1} = \binom{5}{2} = 10 \]

#### Distribution of \(3^1\):
Now, let \(b_1\), \(b_2\), and \(b_3\) represent the powers of 3 in \(x\), \(y\), and \(z\) respectively. We have:
\[ b_1 + b_2 + b_3 = 1 \]
where \(b_1, b_2, b_3 \geq 0\).

Using the "stars and bars" theorem again:
\[ \text{Number of solutions} = \binom{1 + 3 - 1}{3 - 1} = \binom{3}{2} = 3 \]

### Step 3: Total Solutions
To find the total number of ordered pairs \((x, y, z)\), we multiply the number of ways to distribute the factors of 2 and the number of ways to distribute the factors of 3:
\[ \text{Total solutions} = 10 \times 3 = 30 \]

### Conclusion
Thus, the total number of positive integral solutions \((x, y, z)\) such that \(xyz = 24\) is:
\[ \boxed{30} \]

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