A) 3 of 10 is ${10 \choose 3}=\frac{10.9.8}{3.2.1}=10.3.4=120$.
B) 3 of 6 is ${6 \choose 3}=\frac{6.5.4}{3.2.1}=5.4=20$.
C) The probability that a committee is of 3 men is 20/120=1/6.
D) Two men and one woman can be chosen in ${6 \choose 2}{4 \choose 1}=\frac{6.5} {2.1}\frac{4}{1}=15.4=60$.
E) The probability is 60/120=1/2.