| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | To solve this problem, we can break it down into three cases based on the number of red balls selected: 1. **Selecting exactly 2 red balls:** - Choose 2 red balls out of 4: $\binom{4}{2} = 6$ ways. - Choose 2 balls from the remaining 5 (3 green + 2 blue): $\binom{5}{2} = 10$ ways. - Total ways for this case: $6 \times 10 = 60$. 2. **Selecting exactly 3 red balls:** - Choose 3 red balls out of 4: $\binom{4}{3} = 4$ ways. - Choose 1 ball from the remaining 5 (3 green + 2 blue): $\binom{5}{1} = 5$ ways. - Total ways for this case: $4 \times 5 = 20$. 3. **Selecting all 4 red balls:** - There's only 1 way to select all 4 red balls since they are distinct. Adding up the ways from all three cases gives us the total number of ways to choose 4 balls with at least 2 being red: $$60 + 20 + 1 = 81 \text{ ways.}$$ |
81 |
| deepseekr1 | To solve the problem of finding the number of ways to choose 4 balls from a box containing 4 red, 3 green, and 2 blue balls (all distinct) such that at least 2 are red, we can break it down into cases based on the number of red balls selected: 1. **Exactly 2 red balls**: - Choose 2 red balls from 4: \( \binom{4}{2} \) - Choose 2 non-red balls from 5 (3 green + 2 blue): \( \binom{5}{2} \) - Total ways: \( \binom{4}{2} \times \binom{5}{2} = 6 \times 10 = 60 \) 2. **Exactly 3 red balls**: - Choose 3 red balls from 4: \( \binom{4}{3} \) - Choose 1 non-red ball from 5: \( \binom{5}{1} \) - Total ways: \( \binom{4}{3} \times \binom{5}{1} = 4 \times 5 = 20 \) 3. **Exactly 4 red balls**: - Choose 4 red balls from 4: \( \binom{4}{4} \) - Choose 0 non-red balls from 5: \( \binom{5}{0} \) - Total ways: \( \binom{4}{4} \times \binom{5}{0} = 1 \times 1 = 1 \) Summing these cases gives the total number of valid combinations: \[ 60 + 20 + 1 = 81 \] Alternatively, we can calculate the total number of ways to choose 4 balls from 9 and subtract the invalid cases (0 or 1 red balls): - **Total combinations**: \( \binom{9}{4} = 126 \) - **Invalid cases**: - 0 red balls: \( \binom{5}{4} = 5 \) - 1 red ball: \( \binom{4}{1} \times \binom{5}{3} = 4 \times 10 = 40 \) - **Total invalid combinations**: \( 5 + 40 = 45 \) - **Valid combinations**: \( 126 - 45 = 81 \) Both methods confirm the answer is \(\boxed{81}\). |
81 |
| gpto1 | To solve this problem, we need to determine the number of ways to choose 4 balls from a box containing distinct balls of different colors, with the condition that at least 2 of the chosen balls are red. **Given:** - Red balls: 4 (labeled as R₁, R₂, R₃, R₄) - Green balls: 3 (labeled as G₁, G₂, G₃) - Blue balls: 2 (labeled as B₁, B₂) - Total balls: 9 **Objective:** Find the number of ways to choose 4 balls such that at least 2 are red. **Approach:** Since the balls are distinct, even those of the same color, we can use combinations without worrying about identical items. We'll consider all possible cases where the number of red balls in the selection is 2, 3, or 4. **Cases:** 1. **Selecting exactly 2 red balls:** - Choose 2 red balls out of 4: \( C(4, 2) = 6 \) ways. - Choose the remaining 2 balls from the non-red balls (5 balls in total): \( C(5, 2) = 10 \) ways. - Total ways in this case: \( 6 \times 10 = 60 \). 2. **Selecting exactly 3 red balls:** - Choose 3 red balls out of 4: \( C(4, 3) = 4 \) ways. - Choose the remaining 1 ball from the non-red balls: \( C(5, 1) = 5 \) ways. - Total ways in this case: \( 4 \times 5 = 20 \). 3. **Selecting all 4 red balls:** - Choose all 4 red balls: \( C(4, 4) = 1 \) way. - No need to choose any non-red balls. - Total ways in this case: \( 1 \times 1 = 1 \). **Calculations:** - **Case 1:** \( 6 \times 10 = 60 \) ways - **Case 2:** \( 4 \times 5 = 20 \) ways - **Case 3:** \( 1 \times 1 = 1 \) way **Total number of ways:** \( 60 + 20 + 1 = 81 \) ways. **Answer:** **81** |
81 |
| GPT-4o | To determine the number of ways to choose 4 balls from a box containing 4 red, 3 green, and 2 blue balls such that at least 2 of the chosen balls are red, we need to consider different scenarios based on the number of red balls chosen. The scenarios are: 1. Choosing 2 red balls 2. Choosing 3 red balls 3. Choosing 4 red balls ### Scenario 1: Choosing 2 Red Balls - Choose 2 red balls out of 4: \(\binom{4}{2}\) - Choose the remaining 2 balls from the 5 non-red balls (3 green + 2 blue): \(\binom{5}{2}\) Calculating: \[ \binom{4}{2} = \frac{4!}{2!(4-2)!} = \frac{4 \times 3}{2 \times 1} = 6 \] \[ \binom{5}{2} = \frac{5!}{2!(5-2)!} = \frac{5 \times 4}{2 \times 1} = 10 \] Total ways for this scenario: \[ 6 \times 10 = 60 \] ### Scenario 2: Choosing 3 Red Balls - Choose 3 red balls out of 4: \(\binom{4}{3}\) - Choose the remaining 1 ball from the 5 non-red balls: \(\binom{5}{1}\) Calculating: \[ \binom{4}{3} = \frac{4!}{3!(4-3)!} = \frac{4}{1} = 4 \] \[ \binom{5}{1} = \frac{5!}{1!(5-1)!} = \frac{5}{1} = 5 \] Total ways for this scenario: \[ 4 \times 5 = 20 \] ### Scenario 3: Choosing 4 Red Balls - Choose 4 red balls out of 4: \(\binom{4}{4}\) Calculating: \[ \binom{4}{4} = \frac{4!}{4!(4-4)!} = 1 \] Total ways for this scenario: \[ 1 \] ### Total Number of Ways Adding the number of ways from all scenarios: \[ 60 + 20 + 1 = 81 \] Therefore, the total number of ways to choose 4 balls such that at least 2 of them are red is \( \boxed{81} \). |
81 |
| llama3.1-70B-Instruct | Let's break down the problem into cases: **Case 1: Exactly 2 red balls** We choose 2 red balls out of 4 in C(4, 2) = 6 ways. We choose 2 non-red balls out of 5 (3 green + 2 blue) in C(5, 2) = 10 ways. Total ways in this case: 6 × 10 = 60 **Case 2: Exactly 3 red balls** We choose 3 red balls out of 4 in C(4, 3) = 4 ways. We choose 1 non-red ball out of 5 (3 green + 2 blue) in C(5, 1) = 5 ways. Total ways in this case: 4 × 5 = 20 **Case 3: Exactly 4 red balls** We choose all 4 red balls in C(4, 4) = 1 way. No non-red balls are chosen. Total ways in this case: 1 Adding up the ways from all cases, we get: 60 + 20 + 1 = 81 Therefore, there are 81 ways to choose 4 balls, if at least 2 are red. |
81 |
| llama3.1-405B-Instruct | To solve this problem, we can use the concept of combinations. We want to choose 4 balls from a total of 9 balls (4 red, 3 green, and 2 blue), with the condition that at least 2 of the chosen balls are red. Let's break down the problem into cases: 1. Exactly 2 red balls: We can choose 2 red balls from 4 red balls in C(4,2) = 6 ways. We can choose the remaining 2 balls from the 5 non-red balls (3 green and 2 blue) in C(5,2) = 10 ways. So, the total number of ways to choose 4 balls with exactly 2 red balls is 6 x 10 = 60. 2. Exactly 3 red balls: We can choose 3 red balls from 4 red balls in C(4,3) = 4 ways. We can choose the remaining 1 ball from the 5 non-red balls (3 green and 2 blue) in C(5,1) = 5 ways. So, the total number of ways to choose 4 balls with exactly 3 red balls is 4 x 5 = 20. 3. Exactly 4 red balls: We can choose all 4 red balls in C(4,4) = 1 way. Now, we add up the number of ways for each case: 60 (exactly 2 red balls) + 20 (exactly 3 red balls) + 1 (exactly 4 red balls) = 81 Therefore, there are 81 ways to choose 4 balls from the box, with at least 2 of them being red. |
81 |
| DeepSeek-Coder-V2 | To solve the problem of choosing 4 balls from a box containing 4 red, 3 green, and 2 blue balls, with the condition that at least 2 balls must be red, we can break down the problem into different cases based on the number of red balls chosen. ### Case 1: Choosing 2 red balls and 2 other balls - First, choose 2 red balls from the 4 available red balls. The number of ways to do this is given by the combination formula \( \binom{4}{2} \). \[ \binom{4}{2} = \frac{4!}{2!(4-2)!} = \frac{4 \times 3}{2 \times 1} = 6 \] - Next, choose 2 balls from the remaining 5 balls (3 green and 2 blue). The number of ways to do this is given by \( \binom{5}{2} \). \[ \binom{5}{2} = \frac{5!}{2!(5-2)!} = \frac{5 \times 4}{2 \times 1} = 10 \] - Therefore, the total number of ways to choose 2 red balls and 2 other balls is: \[ 6 \times 10 = 60 \] ### Case 2: Choosing 3 red balls and 1 other ball - First, choose 3 red balls from the 4 available red balls. The number of ways to do this is given by \( \binom{4}{3} \). \[ \binom{4}{3} = \frac{4!}{3!(4-3)!} = \frac{4}{1} = 4 \] - Next, choose 1 ball from the remaining 5 balls (3 green and 2 blue). The number of ways to do this is given by \( \binom{5}{1} \). \[ \binom{5}{1} = \frac{5!}{1!(5-1)!} = 5 \] - Therefore, the total number of ways to choose 3 red balls and 1 other ball is: \[ 4 \times 5 = 20 \] ### Case 3: Choosing all 4 red balls - There is only one way to choose all 4 red balls from the 4 available red balls. \[ \binom{4}{4} = 1 \] ### Total Number of Ways Adding up all the cases, we get the total number of ways to choose 4 balls with at least 2 red balls: \[ 60 + 20 + 1 = 81 \] Thus, the number of ways to choose 4 balls from the box with at least 2 red balls is: \[ \boxed{81} \] |
81 |
| Qwen2-72B-Instruct | To solve this problem, we can break it down into three cases based on the number of red balls selected: 1. **Selecting exactly 2 red balls:** - Choose 2 red balls out of 4: $\binom{4}{2} = 6$ ways. - Choose 2 balls from the remaining 5 (3 green + 2 blue): $\binom{5}{2} = 10$ ways. - Total ways for this case: $6 \times 10 = 60$. 2. **Selecting exactly 3 red balls:** - Choose 3 red balls out of 4: $\binom{4}{3} = 4$ ways. - Choose 1 ball from the remaining 5 (3 green + 2 blue): $\binom{5}{1} = 5$ ways. - Total ways for this case: $4 \times 5 = 20$. 3. **Selecting all 4 red balls:** - There's only 1 way to select all 4 red balls since they are distinct. Adding up the ways from all three cases gives us the total number of ways to choose 4 balls with at least 2 being red: $$60 + 20 + 1 = 81 \text{ ways.}$$ |
81 |
| baichuan4 | To find the number of ways to choose 4 balls from a box containing 4 red, 3 green, and 2 blue balls, with at least 2 being red, we can break down the problem into different cases based on the number of red balls chosen and then use combinations to calculate each case. Case 1: Exactly 2 red balls are chosen. If we choose 2 red balls, we need to choose 2 more balls from the remaining 5 non-red balls (3 green and 2 blue). The number of ways to choose 2 red balls out of 4 is C(4, 2), and the number of ways to choose 2 non-red balls out of 5 is C(5, 2). Thus, the total for this case is: C(4, 2) * C(5, 2) Case 2: Exactly 3 red balls are chosen. If we choose 3 red balls, we only need to choose 1 more ball from the remaining 5 non-red balls. The number of ways to choose 3 red balls out of 4 is C(4, 3), and the number of ways to choose 1 non-red ball out of 5 is C(5, 1). Thus, the total for this case is: C(4, 3) * C(5, 1) Case 3: All 4 balls are red. The number of ways to choose all 4 balls as red is simply C(4, 4). Now, let's calculate each combination: C(4, 2) = 4! / (2! * (4-2)!) = 6 C(5, 2) = 5! / (2! * (5-2)!) = 10 C(4, 3) = 4! / (3! * (4-3)!) = 4 C(5, 1) = 5! / (1! * (5-1)!) = 5 C(4, 4) = 4! / (4! * (4-4)!) = 1 Plugging these values into our cases: Case 1: 6 * 10 = 60 Case 2: 4 * 5 = 20 Case 3: 1 Adding up all the cases gives us the total number of ways to choose 4 balls with at least 2 red: 60 (Case 1) + 20 (Case 2) + 1 (Case 3) = 81 So, there are 81 different ways to choose 4 balls from the box with at least 2 being red. |
81 |
| doubao | We can consider three cases: Case 1: Exactly 2 red balls. We choose 2 red balls from 4 in C(4,2) ways. Then we need to choose 2 more balls from the 3 green and 2 blue, which is C(5,2) ways. So the number of ways for this case is C(4,2) * C(5,2). Case 2: exactly 3 red balls. We choose 3 red balls from 4 in C(4,3) ways. Then we choose 1 more ball from the 3 green and 2 blue in C(5,1) ways. So the number of ways for this case is C(4,3) * C(5,1). Case 3: all 4 red balls. This can be done in C(4,4) way. Adding up the number of ways from these three cases gives the total number of ways: C(4,2) * C(5,2) + C(4,3) * C(5,1) + C(4,4) = (6 * 10) + (4 * 5) + 1 = 60 + 20 + 1 = 81 ways. |
81 |
| minimax | To solve this problem, we need to consider the different scenarios in which at least 2 red balls are chosen. We can break this down into three cases: 1. Exactly 2 red balls are chosen. 2. Exactly 3 red balls are chosen. 3. All 4 red balls are chosen. For each case, we will calculate the number of ways to choose the balls and then sum these numbers to get the total number of ways to choose 4 balls with at least 2 being red. Case 1: Exactly 2 red balls are chosen. We have 4 red balls and we need to choose 2, which can be done in C(4,2) ways. Then we need to choose 2 more balls from the remaining 3 green + 2 blue = 5 balls, which can be done in C(5,2) ways. The total number of ways for this case is the product of these two combinations: C(4,2) * C(5,2) Case 2: Exactly 3 red balls are chosen. We have 4 red balls and we need to choose 3, which can be done in C(4,3) ways. Then we need to choose 1 more ball from the remaining 3 green + 2 blue = 5 balls, which can be done in C(5,1) ways. The total number of ways for this case is the product of these two combinations: C(4,3) * C(5,1) Case 3: All 4 red balls are chosen. We have 4 red balls and we need to choose all 4, which can be done in C(4,4) ways. Since we have already chosen all the red balls, we don't need to choose any more balls from the green or blue ones. The total number of ways for this case is simply: C(4,4) Now, let's calculate the combinations: C(4,2) = 4! / (2! * (4-2)!) = (4 * 3) / (2 * 1) = 6 C(5,2) = 5! / (2! * (5-2)!) = (5 * 4) / (2 * 1) = 10 C(4,3) = 4! / (3! * (4-3)!) = 4 / 1 = 4 C(5,1) = 5! / (1! * (5-1)!) = 5 / 1 = 5 C(4,4) = 4! / (4! * (4-4)!) = 1 / 1 = 1 Now we can calculate the total number of ways for each case: Case 1: 6 * 10 = 60 Case 2: 4 * 5 = 20 Case 3: 1 Adding these up gives us the total number of ways to choose 4 balls with at least 2 being red: Total = Case 1 + Case 2 + Case 3 Total = 60 + 20 + 1 Total = 81 Therefore, there are 81 different ways to choose 4 balls from the box with at least 2 being red. |
81 |
| kimi | To solve this problem, we can break it down into cases based on the number of red balls chosen, since we need at least 2 red balls. Case 1: Choose exactly 2 red balls. - There are 4 red balls, and we need to choose 2. This can be done in \( \binom{4}{2} \) ways. - After choosing 2 red balls, we need to choose 2 more balls from the remaining 5 balls (3 green and 2 blue). This can be done in \( \binom{5}{2} \) ways. - The total number of ways for this case is \( \binom{4}{2} \times \binom{5}{2} \). Case 2: Choose exactly 3 red balls. - There are 4 red balls, and we need to choose 3. This can be done in \( \binom{4}{3} \) ways. - After choosing 3 red balls, we need to choose 1 more ball from the remaining 5 balls (3 green and 2 blue). This can be done in \( \binom{5}{1} \) ways. - The total number of ways for this case is \( \binom{4}{3} \times \binom{5}{1} \). Case 3: Choose all 4 red balls. - There are 4 red balls, and we need to choose all 4. This can be done in \( \binom{4}{4} \) ways. - Since we've chosen all the red balls, we don't need to choose any more balls. - The total number of ways for this case is \( \binom{4}{4} \). Now, let's calculate the combinations for each case: Case 1: \( \binom{4}{2} = \frac{4!}{2!(4-2)!} = \frac{4 \times 3}{2 \times 1} = 6 \) \( \binom{5}{2} = \frac{5!}{2!(5-2)!} = \frac{5 \times 4}{2 \times 1} = 10 \) Total for Case 1: \( 6 \times 10 = 60 \) Case 2: \( \binom{4}{3} = \frac{4!}{3!(4-3)!} = \frac{4}{1} = 4 \) \( \binom{5}{1} = \frac{5!}{1!(5-1)!} = 5 \) Total for Case 2: \( 4 \times 5 = 20 \) Case 3: \( \binom{4}{4} = \frac{4!}{4!(4-4)!} = 1 \) Total for Case 3: \( 1 \times 1 = 1 \) Adding up the total number of ways from all cases, we get: Total number of ways = \( 60 + 20 + 1 = 81 \) So, there are 81 ways to choose 4 balls from the box with at least 2 red balls. |
81 |
| wenxinyiyan | 要解决这个问题，我们需要考虑所有可能的选择方式，其中至少有两个红球。我们可以将这些情况分为三种：选择两个红球和两个非红球、选择三个红球和一个非红球、选择四个红球。 1. 选择两个红球和两个非红球： * 选择两个红球的方式有 C(4，2) = 6 种。 * 选择两个非红球的方式有 C(5，2) = 10 种（因为有3个绿球和2个蓝球，共5个非红球）。 * 因此，这种情况下的总选择方式有 6 * 10 = 60 种。 2. 选择三个红球和一个非红球： * 选择三个红球的方式有 C(4，3) = 4 种。 * 选择一个非红球的方式有 C(5，1) = 5 种。 * 因此，这种情况下的总选择方式有 4 * 5 = 20 种。 3. 选择四个红球： * 选择四个红球的方式只有 C(4，4) = 1 种。 现在，我们将这三种情况的选择方式相加，得到总的选择方式：60 + 20 + 1 = 81。 所以，我们可以以81种不同的方式选择4个球，其中至少有两个是红球。 |
81 |
| zhipu | To solve this problem, we need to count the number of ways to choose 4 balls from the box with the condition that at least 2 of them are red. We can approach this by considering different cases based on the number of red balls chosen. The total number of balls is \(4 + 3 + 2 = 9\). ### Case 1: Choosing exactly 2 red balls - Number of ways to choose 2 red balls from 4 red balls: \(\binom{4}{2}\) - Number of ways to choose the remaining 2 balls from the 5 non-red balls (3 green + 2 blue): \(\binom{5}{2}\) So, the number of ways for this case is: \[ \binom{4}{2} \times \binom{5}{2} \] ### Case 2: Choosing exactly 3 red balls - Number of ways to choose 3 red balls from 4 red balls: \(\binom{4}{3}\) - Number of ways to choose the remaining 1 ball from the 5 non-red balls: \(\binom{5}{1}\) So, the number of ways for this case is: \[ \binom{4}{3} \times \binom{5}{1} \] ### Case 3: Choosing exactly 4 red balls - Number of ways to choose 4 red balls from 4 red balls: \(\binom{4}{4}\) - Number of ways to choose 0 balls from the 5 non-red balls: \(\binom{5}{0}\) So, the number of ways for this case is: \[ \binom{4}{4} \times \binom{5}{0} \] Now, we calculate each of these combinations: \[ \binom{4}{2} = \frac{4!}{2!(4-2)!} = \frac{4 \times 3}{2 \times 1} = 6 \] \[ \binom{5}{2} = \frac{5!}{2!(5-2)!} = \frac{5 \times 4}{2 \times 1} = 10 \] \[ \binom{4}{3} = \frac{4!}{3!(4-3)!} = \frac{4}{1} = 4 \] \[ \binom{5}{1} = \frac{5!}{1!(5-1)!} = 5 \] \[ \binom{4}{4} = \frac{4!}{4!(4-4)!} = 1 \] \[ \binom{5}{0} = \frac{5!}{0!(5-0)!} = 1 \] Now, we sum the number of ways from each case: \[ \text{Total ways} = (\binom{4}{2} \times \binom{5}{2}) + (\binom{4}{3} \times \binom{5}{1}) + (\binom{4}{4} \times \binom{5}{0}) \] \[ = (6 \times 10) + (4 \times 5) + (1 \times 1) \] \[ = 60 + 20 + 1 \] \[ = 81 \] Therefore, there are 81 ways to choose 4 balls from the box if at least 2 of them are red. |
81 |
| gpto1 | To solve this problem, we need to determine the number of ways to choose 4 balls from a box containing distinct balls of different colors, with the condition that at least 2 of the chosen balls are red. **Given:** - Red balls: 4 (labeled as R₁, R₂, R₃, R₄) - Green balls: 3 (labeled as G₁, G₂, G₃) - Blue balls: 2 (labeled as B₁, B₂) - Total balls: 9 **Objective:** Find the number of ways to choose 4 balls such that at least 2 are red. **Approach:** Since the balls are distinct, even those of the same color, we can use combinations without worrying about identical items. We'll consider all possible cases where the number of red balls in the selection is 2, 3, or 4. **Cases:** 1. **Selecting exactly 2 red balls:** - Choose 2 red balls out of 4: \( C(4, 2) = 6 \) ways. - Choose the remaining 2 balls from the non-red balls (5 balls in total): \( C(5, 2) = 10 \) ways. - Total ways in this case: \( 6 \times 10 = 60 \). 2. **Selecting exactly 3 red balls:** - Choose 3 red balls out of 4: \( C(4, 3) = 4 \) ways. - Choose the remaining 1 ball from the non-red balls: \( C(5, 1) = 5 \) ways. - Total ways in this case: \( 4 \times 5 = 20 \). 3. **Selecting all 4 red balls:** - Choose all 4 red balls: \( C(4, 4) = 1 \) way. - No need to choose any non-red balls. - Total ways in this case: \( 1 \times 1 = 1 \). **Calculations:** - **Case 1:** \( 6 \times 10 = 60 \) ways - **Case 2:** \( 4 \times 5 = 20 \) ways - **Case 3:** \( 1 \times 1 = 1 \) way **Total number of ways:** \( 60 + 20 + 1 = 81 \) ways. **Answer:** **81** |
81 |
| deepseekv3 | To solve the problem of selecting 4 balls from a box containing 4 red, 3 green, and 2 blue balls, with the condition that at least 2 of the selected balls must be red, we can proceed with the following steps. ### Step 1: Determine the Total Number of Balls First, let's find the total number of distinct balls in the box: - **Red balls:** 4 - **Green balls:** 3 - **Blue balls:** 2 **Total balls = 4 (red) + 3 (green) + 2 (blue) = 9** ### Step 2: Calculate the Total Number of Ways to Choose 4 Balls The total number of ways to choose 4 balls out of 9 without any restrictions is given by the combination formula: \[ \text{Total ways} = \binom{9}{4} = \frac{9!}{4!(9-4)!} = \frac{9!}{4!5!} = 126 \] ### Step 3: Calculate the Number of Ways to Choose 4 Balls with Fewer Than 2 Red Balls The condition requires that we have **at least 2 red balls**. To find this, it's often easier to calculate the total number of ways and subtract the number of ways that do not meet the condition. There are two scenarios where we have fewer than 2 red balls: 1. **0 red balls** 2. **1 red ball** **Case 1: 0 red balls** - We need to choose all 4 balls from the non-red balls (green and blue). - **Number of non-red balls = 3 (green) + 2 (blue) = 5** - Number of ways: \[ \binom{5}{4} = \frac{5!}{4!1!} = 5 \] **Case 2: 1 red ball** - We choose 1 red ball from the 4 available and 3 balls from the non-red balls. - Number of ways: \[ \binom{4}{1} \times \binom{5}{3} = 4 \times 10 = 40 \] **Total ways with fewer than 2 red balls = 5 (Case 1) + 40 (Case 2) = 45** ### Step 4: Calculate the Number of Ways with At Least 2 Red Balls Subtract the number of ways with fewer than 2 red balls from the total number of ways: \[ \text{Valid ways} = \text{Total ways} - \text{Ways with fewer than 2 red balls} = 126 - 45 = 81 \] ### Final Answer The number of ways to choose 4 balls with at least 2 red balls is **81**. |
81 |