One of the easier ways would be to calculate all possible 3 memeber subcomittess from 10 and then subtratc from it the total number of subcomittees possible 3 memeber subcomittess with no french(from the 6 that are spanish and german)

Total number of 3 member subcomitees that can be formed from 10 = 10C3 = 120

Total number of subcomitees that can be formed from Spanish or German = 6C3 = 20

at least 1 member who teaches French = 120 - 20 =100

Choose E)

Thanks!

The other way would be the longer way

Problem says ..at least 1 ..so the options are
a) 1 French + 2 others
1 French = 4
2 others = 6C2 = 15
Total = 15 * 4 = 60

b) 2 F + 1 O
2F = 4C2 = 6
1O = 6
Total = 6*6 = 36

c) 3 F
= 4C3 = 4

So total number of ways = 60 + 36 + 4 = 100