what is the difference between ${10 \choose 2}$ and ${10 \choose 1}\times{9 \choose 1}$? I'm not able to understand the difference between these two.
Don't both just give the number of ways of selecting $2$ objects from a total of $10$?
Maybe the difference is that ${10 \choose 2}$ gives the number of ways of selecting $2$ distinct objects from a total of $10$, but ${10 \choose 1}$ gives the number of ways of selecting $1$ object from a total of $10$, but ${9 \choose 1}$ gives the number of ways of selecting $1$ object from a possible $9$ left options, so shouldn't multiplying ${10 \choose 1}$ with ${9 \choose 1}$ give the same result logicaly?
 A: The difference is the duplication of pairs you get from 10C1 followed by 9C1.
For example, if the set you are choosing from is {a, b, c, d, e, f, g, h, i, j}
Then the choices for 10C1 followed by 9C1 will include (a, b) and (b, a) whereas the choices for 10C2 will include only one of the two.
A: $10C_2$ means selecting $2$ items from a set of $10$ items.
$10C_1\cdot9C_1$ means selecting $1$ item from $10$ and then selecting $1$ item from the rest $9$.
Though you are eventually selecting a total of $2$ items, you should also take a note the parent set from which you are selecting is different in both the cases.
A: One way of seeing the difference is this.
Suppose you are to pick from the set of integers $\{1,2,3,4,5,6,7,8,9,10\}$.
For $\binom{10}{2}$ you choose two elements
"at the same time", that is, picking $2$ and $3$ is the same
as picking $3$ and $2$.
But for $\binom{10}{1}\binom{9}{1}$ the elements are picked in a definite
sequence, that is, picking $2$ the first time and $3$ the second time
is counted separately from picking $3$ the first time and $2$ the second time.
Often $\binom{10}{2}$ is called the number of combinations of $2$
elements you can pick from a set of $10$ elements.
If you are counting permutations of $2$ elements out of $10$
rather than combinations, that is, it matters in what order the elements
are placed rather than just what elements are in the selected subset,
then in fact $\binom{10}{1}\binom{9}{1} = 10\cdot 9$
is exactly the right number of permutations.
A: The number $\binom{10}{2}$ is the number of ways we can select a subset of two elements from a set with ten elements.  Assuming you are making both selections from the same set, the number $\binom{10}{1}\binom{9}{1}$ is the number of ways we can select a subset of one element from a set of ten elements and then choose a subset of one element from the remaining nine elements of the ten element set.  That is, $\binom{10}{1}\binom{9}{1}$ represents the number of ways we can make an ordered selection of two elements from the ten element set. If we are just interested in which elements we have selected, selecting $0$ and then $1$ is the same as selecting $1$ and then $0$.  Using $\binom{10}{1}\binom{9}{1}$ counts them separately, so $$\binom{10}{2} = \frac{1}{2!}\binom{10}{1}\binom{9}{1}$$
because there are $2! = 2$ orders in which we can select the same two elements.
A: No because the product 
$${10 \choose 1}\times{9 \choose 1}$$
does not account for inter-dependencies as the direct calculation (else everything could be reduced, see it as such)
$${10 \choose 2}$$
The inter-dependencies are counted by a convolution-like identity (which is indeed an identity of the binomial coefficients)

For your example the inter-dependencies would be the duplication of pairs you get from ${10 \choose 1}$ followed by ${9 \choose 1}$ (as already mentioned), plus the ordering.
One can easily validate this by direct calculation of the two terms and see if they are equal.
