Show that if $B$ and $C$ are independent then $P[A\mid B\cap C] = \frac{P[A\cap B\mid C]}{P[B]}$ I didn't fit all of the problem in the title, but here it is in full:

a) Given events $A$, $B$ and $C$, show that if $B$ and $C$ are independent then: 
$$P[A\mid B\cap C] = \frac{P[A\cap B\mid C]}{P[B]}$$
  b) In how many ways can ten €1 coins be distributed amongst 3 persons.

I'm a little lost on how to approach this - I understand the difference between mutually exclusive and independent events (being that mutually exclusive events both cannot happen at once, and independent events do not affect one another) but I don't entirely understand what formula I should be using to figure out how to solve the problem.
 A: For (a) 
Given: Events B and C are independent
To prove: $$P[A|B\cap C] = \frac{P[A\cap B|C]}{P[B]}$$
Proof: Events B and C are independent:
$$P[B\cap C] = P(B) * P(C) $$
Thus,
$$P[A|B\cap C] = \frac{P(A\cap B\cap C)} {P(B)\cap (C)}$$
$$P[A|B\cap C] = \frac{P(A\cap B\cap C)} {P(B) * P(C)}$$
$$P[A|B\cap C] = \frac{P[(A\cap B)\cap C]} {P(B) * P(C)}$$
$$P[A|B\cap C] = \frac{P(A\cap B|C)* P(C)} {P(B) * P(C)}$$
$$P[A|B\cap C] = \frac{P(A\cap B|C)} {P(B)}$$
For (b)
Denoting the three persons by A, B and C and a €1 coin by ⨁, then some of the possible ways of distributing the ten coins among three persons A, B and C are:
$$\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\text{A}\qquad\text{B}\qquad\text{C}$$
$$⨁ ⨁ | ⨁ ⨁ ⨁ ⨁ | ⨁ ⨁ ⨁ ⨁ -> 2 \qquad 4 \qquad 4$$
$$⨁ ⨁ ⨁ | ⨁ | ⨁ ⨁ ⨁ ⨁ ⨁ ⨁-> 3 \qquad 1 \qquad 6$$
$$| ⨁ ⨁ ⨁ ⨁ ⨁ ⨁ ⨁ ⨁ ⨁ | ⨁-> 0 \qquad 9 \qquad 1$$
$$⨁ | ⨁ ⨁ ⨁ ⨁ ⨁ ⨁ ⨁ ⨁ | ⨁-> 1 \qquad 8 \qquad 1$$
$$| ⨁ ⨁ ⨁ ⨁ ⨁ ⨁ ⨁ ⨁ ⨁ ⨁ |-> 0 \qquad 10 \qquad 0$$
Thus, the number of ways of distributing the 10 coins amongst the 3 persons is equivalent to the number of ways of arranging 12 objects (10 coins and 2 dividers) among themselves.
$$\therefore \text{Number of ways} = \frac{12!}{10!2!} = \frac{12*11}{2*1}$$
$$\qquad\qquad\qquad\qquad\qquad = \underline{66}$$
Alternative method for (b)
Possible ways of distributing 10 identical coins among 3 persons A, B and C:
$$ A \qquad B \qquad C $$
$$ 0 \qquad 0 \qquad 10 $$
$$ 1 \qquad 1 \qquad 8 $$
$$ 2 \qquad 2 \qquad 6 $$
$$ 3 \qquad 3 \qquad 4 $$
$$ 4 \qquad 4 \qquad 2 $$
$$ 5 \qquad 5 \qquad 0 $$
The above: $$6*\frac {3!}{2!} = 18$$
$$ A \qquad B \qquad C $$
$$ 0 \qquad 1 \qquad 9 $$
$$ 0 \qquad 2 \qquad 8 $$
$$ 0 \qquad 3 \qquad 7 $$
$$ 0 \qquad 4 \qquad 6 $$
$$ 1 \qquad 2 \qquad 7 $$
$$ 1 \qquad 3 \qquad 6 $$
$$ 1 \qquad 4 \qquad 5 $$
$$ 2 \qquad 3 \qquad 5 $$
The above: $$8*3! = 48$$
$$\therefore \text{Number of possible ways of distributing coins} = 18+48 = \underline{66}$$
A: In general we have the rule: $$P(A\mid B)P(B)=P(A\cap B)$$

a)
Since $B$ and $C$ are independent:$$P\left(A\mid B\cap C\right)P\left(B\right)P\left(C\right)=P\left(A\mid B\cap C\right)P\left(B\cap C\right)=P\left(A\cap B\cap C\right)=P\left(A\cap B\mid C\right)P\left(C\right)$$
Now divide both sides with $P\left(B\right)P\left(C\right)$.

b)
To be found is the number of sums $c_{1}+c_{2}+c_{3}=10$ where the
$c_{i}$ are nonnegative integers.
This can be found by application of stars and bars resulting in $\binom{10+2}{2}=66$
solutions.
