A committee of 6 people is to be chosen from 9 women and 5 men. Find the number of ways in which the 6 people can be chosen if there must be more women than men on the committee. ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................
Exam No：9709_w20_qp_53 Year：2020 Question No：3(a)
Answer:
Scenarios:
\(6 \mathrm{~W} 0 \mathrm{M}^{9} \mathrm{C}_{6}=84\)
\(5 \mathrm{~W} 1 \mathrm{M}^{9} \mathrm{C}_{5} \times{ }^{5} \mathrm{C}_{1}=126 \times 5=630\)
\(4 \mathrm{~W} 2 \mathrm{M}^{9} \mathrm{C}_{4} \times{ }^{5} \mathrm{C}_{2}=126 \times 10=1260\)
Total \(=1974\)
\(6 \mathrm{~W} 0 \mathrm{M}^{9} \mathrm{C}_{6}=84\)
\(5 \mathrm{~W} 1 \mathrm{M}^{9} \mathrm{C}_{5} \times{ }^{5} \mathrm{C}_{1}=126 \times 5=630\)
\(4 \mathrm{~W} 2 \mathrm{M}^{9} \mathrm{C}_{4} \times{ }^{5} \mathrm{C}_{2}=126 \times 10=1260\)
Total \(=1974\)
Knowledge points:
5.2.1 understand the terms permutation and combination, and solve simple problems involving selections
Solution:
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