A class has divided 6 students randomly into teams A and B. What is the probability that 3 students from team A will come first, second and third? My take on the problem is, considering the players to be indistinguishable individually other than by their teams i.e. the players are A,B,A,B,A,B. They can be arranged in $\frac{6!}{3!3!}$ ways.
Assuming the arrangement of 3 players of team A becoming 1st, 2nd, 3rd to be AAABBB, this arrangement can be only done in 1 way the answer to the question comes out to be
$$\frac{1}{\frac{6!}{3!3!}}= \frac{3!3!}{6!}$$.
but the options given for this question are
a) $\frac{3!}{6!}$
b) $\frac{1}{6!}$
c) $\frac{3!2}{6!}$
d) $\frac{3}{6!}$
Help me understand where I'm going wrong.
 A: The probability that the first student that walks in will be from team A is: $\frac{3}{6}$
Given that the first student to walk in was from team A, the probability that the second student that walks in will be from team A is: $\frac{2}{5}$
Given that the first two students to walk in were from team A, the probability that the third student that walks in will be from team A is: $\frac{1}{4}$
The multiplication rule then gives us that the probability that the first three students to walk in will all be from team A is:
$$(\frac{3}{6})(\frac{2}{5})(\frac{1}{4})=\frac{3!3!}{6!}$$
A: Combinatorially:
There are ${6\choose 3}=20$ ways to choose the top three from the 6 students.  There is 1 way that all three of the top three can come from group A.
The probability then is $\frac{1}{20}$=0.05.
A: I was thinking of the case where all the players are distinguishable : $A_1,A_2,A_3,B_1,B_2,B_3$ then we need to find the probability of the arrangement $AAAB_iB_jB_k$. Number of such cases are $3!$ and the total number of arrangements of $A_1,A_2,A_3,B_1,B_2,B_3$ are $6!$.
So the required probability becomes $\frac{3!}{6!}$.
Although I'm unsure whether this is the right wat to go about the problem.
