Rearrangement of Students (flaw in my solution) 
There are 11 students in a class including A, B and C. The 11 students have to form a straight line. Provided that A cannot be the first person in the line, what is the probability that in any random rearrangement of line, A comes before B and C.

For eg, this is a valid rearrangement (1-8 are other students)
1 2 3 A 4 5 C 6 7 B 8
Here's my solution -
Probability that A goes before B and C without any restrictions is $\frac13$. (Notice the symmetry. The answer will be same for B goes first and C goes first.)
Probability that A goes first without any restrictions is $\frac{1}{11}$
Hence the answer is
$$\frac13 - \frac{1}{11} = \frac{8}{33}$$
But the answer is $\frac{4}{15}$ according to my textbook. Please help me to find flaw in my solution.
 A: See total ways of arranging anyhow is $10.10!$. Now the constrained ways .  As a A cannot be first so lets place A at 2 nd place so B,C has to come after let it be CB or BC it doesnt matter so we need to multiply all terms with $2!$ now as A is 2nd we have ${9\choose 2}.2!$ options for B,C and $8!$ for remaining now if A is at $3$ place then we have ${8\choose 2}.2!$ for B,C and $8!$ for others so this can be done for A till $9th$ place . i hope you know the reason . so our probability becomes $$\frac{(2!.8!.({9\choose 2}+{8\choose 2}+...+{2\choose 2})}{10.10!}=\frac{4}{15}$$
A: $\binom{10}{3}$ is the number of ways to select 3 slots in 10 possiblee. Now just account for the fact that b and c can interchange for each choice of slots.
EDIT: Once you have done that, multiply by $8!$ (rearrangement of the remaining students) and divide through $10!10$ (total number of arrangements)
A: When the problem says "Provided $A$ cannot be the first person in line, what is the probability..." it's actually asking for the conditional probability that $A$ comes before $B$ and $C$ given that $A$ is not the first person in line.  So if we let $E$ be the event that $A$ comes before $B$ and $C$, and $F$ be the event that $A$ is not the first person in line, what we want is
$$P(E|F)={P(E\cap F)\over P(F)}$$
Now what you computed in ${1\over3}-{1\over11}={8\over33}$ is actually the numerator, $P(E\cap F)$.  For the probability the problem calls for, it remains to note that $P(F)={10\over11}$, so that
$$P(A\text{ comes before $B$ and $C$ }|\,A\text{ is not first in line})={8/33\over10/11}={4\over15}$$
A: It is sure that possibility A comes before B,C is 1/3. But A must not be the first.
$○○CB$
possible A is 1　pattern
$○○○CB$
2 pattern
・
・
・
8 patterns
1+2+・・・8/1+2+3+・・+9
=55-19/55-10　　
=36/45
=4/5
And then
1/3*4/5=4/15
A: Short way:
As OP has remarked, by symmetry, the probability that $A$ comes before $B$ and $C$ is $\dfrac13$.
The only object that need concern us is the one immediately preceding $A$.
(Others won't affect the probability computation)
There are $8$ ways with the constraints, as against $10$ unconstrained ways for this object.
Thus $Pr = \dfrac13\cdot\dfrac8{10} = \dfrac4{15}$

Added explanation:
For constrained arrangements, we know that there are $8$ ways to choose the ball *immediately preceding $A$, both lumped together as a $\huge\bullet$
$B$ (say) can be put next to it in one way: ${\huge\bullet}\Large\uparrow\bullet\uparrow$, but C can now be introduced in $2$ ways.
The next object can now be introduced in four ways   $\Large\uparrow{\huge\bullet}\Large\uparrow\bullet\uparrow\bullet\uparrow$, including before $A$, and for each  succeeding object, the number of ways will increase by one.
Similarly for unconstrained ways, except that it will start with $10$ choices for the object immediately preceding $A$, and succeeding objects can be placed on either side.
We thus get $\dfrac {8\times1\times2\times4\times 5\times ...\times 10}{10\times2\times3\times4\times 5\times ...\times 10}$
Apart from the initial $\dfrac8{10}$, the rest of it simplifies to $\dfrac13$,
the symmetric probability that $A$ comes before $B$ and $C$
So we can clearly see the rationale of the short way
A: There are two possibilities for the places for A, B, C:
1) If they do not include the first place, then there are $\dbinom{10}{3}$ choices for the combined places for A, B, C, and only $\frac{1}{3}$ of the arrangements will have A before B and C.
2) If they do include the first place, then there are $\dbinom{10}{2}$ choices for the combined positions for A, B, C, and only $\frac{2}{3}$ of the arrangements will not have A in the first place.
Therefore the probability is given by $\displaystyle\frac{\frac{1}{3}\binom{10}{3}}{\binom{10}{3}+\frac{2}{3}\binom{10}{2}}=\frac{40}{120+30}=\frac{4}{15}$.
