Probability Doubt about selection A bus contain 10 tickets 5 printed with 'I' and 5 printed with 'T', 3 tickets are drawn without Replacement and arrange in the same Order in which they are drawn on the table. Find the Probability that IIT is formed.
1st approach:
$\frac{5}{10} \times\frac{4}{9}\times\frac{5}{8}= \frac{5}{36}$
2nd Approach:
$\left (\binom{5}{2}\times\binom{5}{1}  \right )/ \binom{10}{3}$ which is equal to 5/12
My doubt is why they aren't equal can you please explain the concept I missed in this question?
 A: The first approach is exactly the probability that the first is I, then the second is I, then the third is T. So this is correct.
The second probability has a denominator of $\binom{10}3$, which is the number of different sets of three tickets you can pull out, and a numerator of $\binom 52\binom 51$, which is the number of ways to choose a set of two Is and one T. That is, if we mark the tickets with numbers to make them distinct, the denominator counts things like "the first three tickets are the set $\{T_1,T_4,I_2\}$"; notice that this is the same as "the first three tickets are the set $\{I_2,T_4,T_1\}$".
So this doesn't give the right answer, since it only considers the first three tickets as a set, and doesn't take order into account. This actually counts the probability that there are two Is and a T in any order; since this covers three possible cases IIT, ITI and TII, you need to divide by 3 to get the right answer - and $$\frac13\times\frac{\binom 52\binom 51}{\binom{10}3}$$ simplifies to match the first answer.
A: First one is correct and second not, because you did not count  the arrangment. 
Calculation in the second way also include ITI or TII to be formed and that is what you don't want, so it is incorrect.
