Number of positive unequal integral solution of the equationi `x+y+z=6` is

Options

A. 4!
B. 3!
C. 5!
D. `2xx4!`

Answer

To find the number of positive unequal integral solutions of the equation \( x + y + z = 6 \), we will follow these steps:

### Step 1: Understand the problem
We need to find the number of positive integral solutions for the equation \( x + y + z = 6 \) where \( x, y, z \) are all different (unequal).

### Step 2: Transform the variables
Since \( x, y, z \) must be positive integers, we can redefine them as:
- Let \( x = a + 1 \)
- Let \( y = b + 1 \)
- Let \( z = c + 1 \)

where \( a, b, c \) are non-negative integers. This transformation ensures that \( x, y, z \) are at least 1.

### Step 3: Substitute into the equation
Substituting these into the original equation gives:
\[ (a + 1) + (b + 1) + (c + 1) = 6 \]
which simplifies to:
\[ a + b + c = 3 \]

### Step 4: Find the total non-negative integral solutions
The number of non-negative integral solutions to the equation \( a + b + c = 3 \) can be found using the "stars and bars" theorem. The formula for the number of solutions is given by:
\[ \text{Number of solutions} = \binom{n + k - 1}{k - 1} \]
where \( n \) is the sum (3 in this case) and \( k \) is the number of variables (3 here).

So we have:
\[ \text{Number of solutions} = \binom{3 + 3 - 1}{3 - 1} = \binom{5}{2} = \frac{5 \times 4}{2 \times 1} = 10 \]

### Step 5: Adjust for unequal solutions
Now, we need to consider that \( x, y, z \) must be unequal. The total solutions we found (10) include cases where some of the variables may be equal.

To find the cases where at least two variables are equal, we can use the following reasoning:
1. Assume two variables are equal, say \( x = y \). Then we have:
\[ 2x + z = 6 \implies z = 6 - 2x \]
The possible values for \( x \) (and hence \( y \)) are 1, 2, or 3, leading to the pairs:
- \( (1, 1, 4) \)
- \( (2, 2, 2) \)
- \( (3, 3, 0) \) (not valid since \( z \) must be positive)

Thus, we have the valid pairs: \( (1, 1, 4) \) and \( (2, 2, 2) \) which are not valid as they are not unequal.

2. Therefore, we have only one case where two variables are equal, which is \( (1, 1, 4) \).

### Step 6: Calculate the final count of unequal solutions
From the total of 10 solutions, we subtract the invalid cases:
- Total valid solutions = \( 10 - 1 = 9 \)

However, we need to ensure that we only count distinct arrangements of the valid solutions. The valid arrangements of \( (1, 2, 3) \) can be permuted in \( 3! = 6 \) ways.

### Conclusion
Thus, the number of positive unequal integral solutions to the equation \( x + y + z = 6 \) is \( 6 \).