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 at most three of them are red is _________

Answer

To solve the problem of finding the number of ways to draw 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 can break it down into cases based on the number of red marbles drawn.

### Step-by-Step Solution:

1. **Identify the total number of marbles:**
- Red marbles = 5
- Black marbles = 4
- White marbles = 3
- Total marbles = 5 + 4 + 3 = 12

2. **Define the cases:**
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 select all 4 marbles from the black and white marbles.
- Total black and white marbles = 4 + 3 = 7
- Number of ways = \( \binom{7}{4} \)

**Case 2: 1 red marble**
- We select 1 red marble and 3 from the remaining black and white marbles.
- Number of ways = \( \binom{5}{1} \times \binom{7}{3} \)

**Case 3: 2 red marbles**
- We select 2 red marbles and 2 from the remaining black and white marbles.
- Number of ways = \( \binom{5}{2} \times \binom{7}{2} \)

**Case 4: 3 red marbles**
- We select 3 red marbles and 1 from the remaining black and white marbles.
- Number of ways = \( \binom{5}{3} \times \binom{7}{1} \)

4. **Calculate each binomial coefficient:**
- \( \binom{7}{4} = \frac{7!}{4!(7-4)!} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35 \)
- \( \binom{5}{1} = 5 \)
- \( \binom{7}{3} = \frac{7!}{3!(7-3)!} = \frac{7 \times 6}{2 \times 1} = 21 \)
- \( \binom{5}{2} = \frac{5!}{2!(5-2)!} = \frac{5 \times 4}{2 \times 1} = 10 \)
- \( \binom{7}{2} = \frac{7!}{2!(7-2)!} = \frac{7 \times 6}{2 \times 1} = 21 \)
- \( \binom{5}{3} = \frac{5!}{3!(5-3)!} = \frac{5 \times 4}{2 \times 1} = 10 \)
- \( \binom{7}{1} = 7 \)

5. **Substituting the values back:**
- Case 1: \( 35 \)
- Case 2: \( 5 \times 21 = 105 \)
- Case 3: \( 10 \times 21 = 210 \)
- Case 4: \( 10 \times 7 = 70 \)

6. **Total number of ways:**
- Total = Case 1 + Case 2 + Case 3 + Case 4
- Total = \( 35 + 105 + 210 + 70 = 420 \)

Thus, the final answer is **420**.