In how many ways can a team of 4 people be chosen from 10 people if 2 of the people, Ross and Lionel, refuse to be in the team together?
Solution
To solve this problem, we need to consider the restriction that Ross (R) and Lionel (L) cannot be in the team together.
Method 1: Consider the cases separately:
- Choose R but not L: Choose 3 more people from the remaining 8, which is \\(^8C_3 = 56\\) ways.
- Choose L but not R: Choose 3 more people from the remaining 8, which is \\(^8C_3 = 56\\) ways.
- Choose neither R nor L: Choose all 4 people from the remaining 8, which is \\(^8C_4 = 70\\) ways.
Summing these scenarios gives: \\((56 + 56 + 70 = 182)\\) ways.
Method 2: Use the complement principle:
- Calculate the total number of ways to choose 4 people from 10 without restriction: \\(^10C_4 = 210\\) ways.
- Subtract the number of ways to choose both R and L in the team: Choose 2 more people from the remaining 8, which is \\(^8C_2 = 28\\) ways.
Thus, the number of valid ways is \\(210 - 28 = 182\\) ways.
Method 3: Consider excluding R or L:
- Exclude R: Choose 4 people from the remaining 9, which is \\(^9C_4 = 126\\) ways.
- Exclude L: Choose 4 people from the remaining 9, which is \\(^9C_4 = 126\\) ways.
- Exclude both R and L: Choose 4 people from the remaining 8, which is \\(^8C_4 = 70\\) ways.
Summing the first two scenarios and subtracting the third gives: \\(126 + 126 - 70 = 182\\) ways.
All methods confirm that the number of ways is 182.
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