The number of solutions of the equation `x+y+z =10` in positive integers `x,y,z` is equal to

Options

A. `36`
B. `55`
C. `72`
D. `45`

Answer

To find the number of solutions of the equation \( x + y + z = 10 \) in positive integers \( x, y, z \), we can follow these steps:

### Step 1: Understand the problem
We need to find the number of positive integer solutions for the equation \( x + y + z = 10 \). Since \( x, y, z \) must be positive integers, we cannot have any of them equal to zero.

### Step 2: Transform the variables
To convert the problem into one that allows for zero values, we can make a substitution. Let:
- \( x = a + 1 \)
- \( y = b + 1 \)
- \( z = c + 1 \)

where \( a, b, c \) are non-negative integers (i.e., they can be zero or more). This transformation ensures that \( x, y, z \) are always positive.

### Step 3: Rewrite the equation
Substituting the new variables into the original equation gives:
\[ (a + 1) + (b + 1) + (c + 1) = 10 \]
This simplifies to:
\[ a + b + c + 3 = 10 \]
or
\[ a + b + c = 7 \]

### Step 4: Apply the stars and bars theorem
Now, we need to find the number of non-negative integer solutions to the equation \( a + b + c = 7 \). We can use the "stars and bars" theorem, which states that the number of ways to distribute \( n \) indistinguishable objects (stars) into \( r \) distinguishable boxes (variables) is given by:
\[ \binom{n + r - 1}{r - 1} \]
In our case:
- \( n = 7 \) (the total we want)
- \( r = 3 \) (the number of variables: \( a, b, c \))

### Step 5: Calculate the number of solutions
Using the formula:
\[ \text{Number of solutions} = \binom{7 + 3 - 1}{3 - 1} = \binom{9}{2} \]

### Step 6: Compute \( \binom{9}{2} \)
Calculating \( \binom{9}{2} \):
\[ \binom{9}{2} = \frac{9!}{2!(9-2)!} = \frac{9 \times 8}{2 \times 1} = \frac{72}{2} = 36 \]

### Final Answer
Thus, the number of solutions of the equation \( x + y + z = 10 \) in positive integers is **36**.

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