a. 10C3=10!/3!(10-3)!= 362880/30240=120 b. 4C2x6C2=4!/2!x6!/2!= 4320 c. 6C3=6!/3!(6-3)!=720/36=20 d. 6C3/10C3=6!/3!/10!/3!= 120/604800 This is the work that I have done so far
(a) Correct. You can choose make your first choice 10 ways. For each of those, you can make your second choice 9 ways, and for each of those 90, you can make your third choice 8 ways. That makes 720 ways that you can choose.
\(\displaystyle 720 = 10 * 9 * 8 = \dfrac{10!}{7!}.\)
But that counts Alice first, Bob second, and Carol third as different from Bob first, Alice second, and Carol third, and we do not care in what order they were selected.
ABC
ACB
BAC
BCA
CAB
ABC
6 duplicates, but 6 = 3 * 2 * 1
\(\displaystyle 120 = \dfrac{720}{6} = \dfrac{10!}{7! * 3!} = \dfrac{10!}{(10 - 3)! * 3!} = \dbinom{10}{3}.\)
Perfect. Well done.
(b) What happened here. How many combinations of boys can you have?
You have six boys to choose from so you have 6 ways to make your first choice and 5 ways to make your second choice for each of your first choices. But once again, we do not care which was chosen first and which second.
Dave and Ed and Ed and David; it's all the same to us.
\(\displaystyle 15 = \dfrac{30}{2} = \dfrac{6 * 5}{2} = \dfrac{6!}{4! * 2!} = \dfrac{6!}{(6 - 2)! * 2!} = \dbinom{6}{2}.\)
Now recalculate the number of prossible combinations of girls.
