A dog breeder currently has 9 breeding dogs. 6 of the dogs have exactl

Please help me with a fundamental understanding:

I solved this Q using the reverse technique i.e.:
Probability of not getting any littermates = prob of not getting (a. any of littermate doubles + b. any two of littermate triplets)
= (2/9)*(7*8) + (2/9)*(7*8) + (2/9)*(7*8) + (3/9)*(6*8)
= 7/12 + 1/4
= 5/6

So, the below method should work as well:
= (2C1*7C1) / 9C2 + (3C1*6C1) / 9C2

But, it does not, I guess there is some problem with 9C2, because if I replace the same with 9*8, the equation becomes the same as above

I am not able to point out the fundamental flaw in 9C2 method as we use it quite often in similar Qs

For eg:
A 10-member student leadership committee consists of 4 juniors and 6 seniors. Exactly 6 students will be selected from this group to attend a national convention. What is the probability that at least 3 seniors are selected for the committee?

Sol:
= 1- (4C4*6C2) / 10C6
= 13/14


Kindly help me to clear this fundamental flaw in my understanding. I have been banging my head over this for few hours:)