There are ten things . Two things are the same and rest are all different. In how many ways sets of 5 things can be formed from this? 
There are ten things . Two things are the same and rest are all different.  In how many ways sets of 5 things can be formed from this?

I label the elements $\{1,2,3,4,5,6,7,8,9,9\}$ ;  we have two 9 , rest are different. 
(1) Sets of 5 things where no 9 is present can be formed in $\binom{8}{5}$ ways
(2) Sets of 5 things where only one 9 is present can be formed in $2 \binom{8}{4}$ ways
(3) Sets of 5 things where both 9 are present can be formed in $\binom{8}{3}$ ways
So finally we have a total of $\binom{8}{5} + 2 \binom{8}{4} + \binom{8}{3}$
Is it correct?
 A: Your approach of distinguishing between cases when no $9$ is present, only one $9$ is present, and two $9$'s are present is sound.  You handled the cases when no $9$ is present and two $9$'s are present correctly.  However, you treated the two $9$'s as if they were distinguishable in the case when only one $9$ is present.  Since they are not distinguishable, the set is determined by which four of the other eight elements are selected with a $9$.  Hence, the number of five-element sets that contain a single $9$ is $$\binom{8}{4}$$ and the total number of five-element sets (really, multisets) that can be formed is 
$$\binom{8}{5} + \binom{8}{4} + \binom{8}{3}$$
A set is determined by the elements it contains.  The number of times that they are listed does not matter.  Hence, the set 
$$\{1, 2, 3, 9, 9\} = \{1, 2, 3, 9\}$$
Therefore, from the point of view of set theory, you can have at most one $9$ in a set.  If that was the intention, the number of five-element sets that can be formed is 
$$\binom{8}{4} + \binom{8}{5}$$
since we could have either no $9$'s or exactly one $9$ in the set.  The wording suggests that the number of five-element multisets that can be selected from a ten-element multiset was intended.
A: I think your approach is right. But the second case ans is wrong. As we have two objects which are actully the same so it doesnt matter whether you take first 9 or the second 9 of your list. So the number of ways of choosing a set of 5 elements including only one 9, out of the given 10 elements will be $C^8_4$. So the ans is $C^8_5+C^8_4+C^8_3$
A: The number of such (multi)sets are the number of $5$-subsets in a $10$-set, $\binom{10}{5}$, minus number of $4$-subsets in a $8$-set,  $\binom{8}{4}$ (because the number of $5$-subsets with one 9 were counted twice).
Hence, we obtain
$$\binom{10}{5}-\binom{8}{4}=182.$$
You will obtain the same number as soon as you correct your case (2).
