The remainder when 1!+2!+3!+.....95! is divided by 15 is - (a) 3 (b) 5 (c) 8 (d) 12

We are asked to find the remainder when \(1! + 2! + 3! + \dots + 95!\) is divided by 15.

### Step-by-step Solution:

1. **Factorial and divisibility:**
- The factorial of any number greater than or equal to 5 includes the factor 15. This is because:
\[
5! = 5 \times 4 \times 3 \times 2 \times 1 = 120
\]
Since 120 is divisible by 15, all factorials from \(5!\) onwards will be divisible by 15. Thus, they will not contribute to the remainder when divided by 15.

2. **Simplifying the sum:**
- The sum can be reduced to:
\[
1! + 2! + 3! + 4!
\]
because all terms from \(5!\) onwards are divisible by 15 and contribute 0 to the remainder.

3. **Calculate the factorials:**
- \(1! = 1\)
- \(2! = 2\)
- \(3! = 6\)
- \(4! = 24\)

4. **Add these factorials:**
\[
1! + 2! + 3! + 4! = 1 + 2 + 6 + 24 = 33
\]

5. **Find the remainder when 33 is divided by 15:**
\[
33 \div 15 = 2 \text{ remainder } 3
\]
So, the remainder when \(33\) is divided by \(15\) is 3.

### Final Answer:
The remainder when \(1! + 2! + 3! + \dots + 95!\) is divided by 15 is **3**.

Thus, the correct answer is **(a) 3**.