Solution:
Fathers (F)=5, Teachers (T)=6, students (S)=4
for, T+F+S=7,T≥0, thus F≥1,S≥1 total combination (T,F,S)
(4,1,2),(4,2,1),(5,1,1),(3,2,2)
Case I(T=4,F=1,S=2)
‌=‌6C4⋅‌5C1⋅‌4C2
‌=15×5×6=450
Case II ( T=4,F=2,S=1 )
‌=‌6C4⋅‌5C2⋅‌4C1
‌=15×10×4=600
Case III ( T=5,F=1,S=1 )
‌=‌6C5⋅‌5C1⋅‌4C1
‌=6×5×4=120
Case IV ( T=3,F=2,S=2 )
‌=‌6C3⋅‌5C2⋅‌4C2
‌=20×10×6=1200
Hence, total number of ways
‌=450+600+120+1200
‌=2370
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