An urn contains 5 red marbels,4 black marbels and 3 white marbles. Then the number of ways in which 4 marbles can be drawn so that most three them are red is _________

Answer

To solve the problem of drawing 4 marbles from an urn containing 5 red marbles, 4 black marbles, and 3 white marbles, with the condition that at most 3 of them can be red, we will break down the solution into cases based on the number of red marbles drawn.

### Step-by-Step Solution:

1. **Identify the Total Number of Marbles**:
- Total red marbles = 5
- Total black marbles = 4
- Total white marbles = 3
- Total marbles = 5 + 4 + 3 = 12

2. **Define Cases Based on Number of Red Marbles**:
We need to consider the following cases for drawing 4 marbles:
- Case 1: 0 red marbles
- Case 2: 1 red marble
- Case 3: 2 red marbles
- Case 4: 3 red marbles

3. **Calculate the Number of Ways for Each Case**:

**Case 1: 0 Red Marbles**
- We need to choose all 4 marbles from the remaining 7 (4 black + 3 white).
- Number of ways = \( \binom{5}{0} \times \binom{7}{4} = 1 \times 35 = 35 \)

**Case 2: 1 Red Marble**
- Choose 1 red marble from 5 and 3 from the remaining 7.
- Number of ways = \( \binom{5}{1} \times \binom{7}{3} = 5 \times 35 = 175 \)

**Case 3: 2 Red Marbles**
- Choose 2 red marbles from 5 and 2 from the remaining 7.
- Number of ways = \( \binom{5}{2} \times \binom{7}{2} = 10 \times 21 = 210 \)

**Case 4: 3 Red Marbles**
- Choose 3 red marbles from 5 and 1 from the remaining 7.
- Number of ways = \( \binom{5}{3} \times \binom{7}{1} = 10 \times 7 = 70 \)

4. **Sum the Number of Ways from All Cases**:
- Total ways = Case 1 + Case 2 + Case 3 + Case 4
- Total ways = 35 + 175 + 210 + 70 = 490

### Final Answer:
The total number of ways to draw 4 marbles such that at most 3 of them are red is **490**.