Let N be the number of integral solution of the equation `x+y+z+w=15" where " x ge 0, y gt 5, z ge 2 and w ge 1`. Find the unit digit of N. `

Answer

To solve the problem of finding the number of integral solutions of the equation \(x + y + z + w = 15\) under the constraints \(x \geq 0\), \(y > 5\), \(z \geq 2\), and \(w \geq 1\), we can follow these steps:

### Step 1: Transform the variables
We need to transform the variables to simplify the constraints:
- Let \(y = 6 + t\) where \(t \geq 0\) (since \(y > 5\))
- Let \(z = 2 + s\) where \(s \geq 0\) (since \(z \geq 2\))
- Let \(w = 1 + r\) where \(r \geq 0\) (since \(w \geq 1\))

### Step 2: Substitute into the equation
Substituting these transformations into the original equation:
\[ x + (6 + t) + (2 + s) + (1 + r) = 15 \]
This simplifies to:
\[ x + t + s + r + 9 = 15 \]
Thus, we can rewrite it as:
\[ x + t + s + r = 15 - 9 \]
which simplifies to:
\[ x + t + s + r = 6 \]

### Step 3: Count the non-negative integer solutions
Now we need to find the number of non-negative integer solutions to the equation \(x + t + s + r = 6\). The number of solutions can be found using the "stars and bars" theorem, which states that the number of ways to distribute \(n\) indistinguishable objects (stars) into \(k\) distinguishable boxes (variables) is given by:
\[ \binom{n + k - 1}{k - 1} \]
In our case, \(n = 6\) (the total we need to achieve) and \(k = 4\) (the number of variables: \(x, t, s, r\)). Therefore, we have:
\[ \text{Number of solutions} = \binom{6 + 4 - 1}{4 - 1} = \binom{9}{3} \]

### Step 4: Calculate \(\binom{9}{3}\)
Calculating \(\binom{9}{3}\):
\[ \binom{9}{3} = \frac{9!}{3!(9-3)!} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = \frac{504}{6} = 84 \]

### Step 5: Find the unit digit of \(N\)
Now that we have \(N = 84\), we need to find the unit digit of \(N\). The unit digit of \(84\) is \(4\).

### Final Answer
Thus, the unit digit of \(N\) is \(4\).

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