An urn contains 5 different red and 6 different green balls. In how many ways can 6 balls be selected so that there are atleast two balls of each colour?

Options

A. 425
B. 245
C. 125
D. 625

Answer

To solve the problem of selecting 6 balls from an urn containing 5 different red balls and 6 different green balls, with the condition that there are at least 2 balls of each color, we can break down the selection into different cases based on the number of red and green balls chosen.

### Step 1: Define the Cases
We need to consider the following cases for selecting the balls:
1. 2 red balls and 4 green balls
2. 3 red balls and 3 green balls
3. 4 red balls and 2 green balls

### Step 2: Calculate Each Case

#### Case 1: 2 Red Balls and 4 Green Balls
- The number of ways to choose 2 red balls from 5 is given by the combination formula \( \binom{n}{r} \), where \( n \) is the total number of items to choose from, and \( r \) is the number of items to choose.
- Thus, the number of ways to choose 2 red balls is \( \binom{5}{2} \).
- The number of ways to choose 4 green balls from 6 is \( \binom{6}{4} \).

Calculating these:
\[ \binom{5}{2} = \frac{5!}{2!(5-2)!} = \frac{5 \times 4}{2 \times 1} = 10 \]
\[ \binom{6}{4} = \binom{6}{2} = \frac{6!}{4!(6-4)!} = \frac{6 \times 5}{2 \times 1} = 15 \]
Thus, the total ways for this case is:
\[ 10 \times 15 = 150 \]

#### Case 2: 3 Red Balls and 3 Green Balls
- The number of ways to choose 3 red balls from 5 is \( \binom{5}{3} \).
- The number of ways to choose 3 green balls from 6 is \( \binom{6}{3} \).

Calculating these:
\[ \binom{5}{3} = \frac{5!}{3!(5-3)!} = \frac{5 \times 4}{2 \times 1} = 10 \]
\[ \binom{6}{3} = \frac{6!}{3!(6-3)!} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20 \]
Thus, the total ways for this case is:
\[ 10 \times 20 = 200 \]

#### Case 3: 4 Red Balls and 2 Green Balls
- The number of ways to choose 4 red balls from 5 is \( \binom{5}{4} \).
- The number of ways to choose 2 green balls from 6 is \( \binom{6}{2} \).

Calculating these:
\[ \binom{5}{4} = 5 \]
\[ \binom{6}{2} = \frac{6!}{2!(6-2)!} = \frac{6 \times 5}{2 \times 1} = 15 \]
Thus, the total ways for this case is:
\[ 5 \times 15 = 75 \]

### Step 3: Total Ways
Now, we add the total ways from all three cases:
\[ 150 + 200 + 75 = 425 \]

### Final Answer
The total number of ways to select 6 balls such that there are at least 2 balls of each color is **425**.

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