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Nov 2015 p63 q5
2843
There are 7 Chinese, 6 European and 4 American students at an international conference. Four of the students are to be chosen to take part in a television broadcast. Find the number of different ways the students can be chosen if at least one Chinese and at least one European student are included.
Solution
To solve this problem, we need to consider the different combinations of students that include at least one Chinese and one European student. We will use combinations to calculate the number of ways to choose students.
Let C, E, and A represent the number of Chinese, European, and American students chosen, respectively. We have the following cases:
C = 1, E = 1, A = 2:
\(7 \times 6 \times \binom{4}{2} = 252\)
C = 1, E = 2, A = 1:
\(7 \times \binom{6}{2} \times 4 = 420\)
C = 1, E = 3, A = 0:
\(7 \times \binom{6}{3} \times 1 = 140\)
C = 2, E = 1, A = 1:
\(\binom{7}{2} \times 6 \times 4 = 504\)
C = 2, E = 2, A = 0:
\(\binom{7}{2} \times \binom{6}{2} \times 1 = 315\)
C = 3, E = 1, A = 0:
\(\binom{7}{3} \times 6 \times 1 = 210\)
Adding these cases together gives the total number of ways:
