Find the probability of drawing a black and a white ball

If $3$ balls are drawn from a bowl containing $6$ white and $5$ black balls, what is the probability that one of the balls is white and the other two black?

Answer: $$\frac { \binom {6}{1} \binom {5}{2} }{ \binom {11}{3} } = \frac {4}{11}$$

Now, my question is, why isn't it

$$\frac { \binom {6}{1} \binom {5}{2} } { \frac { \binom {11}{3}}{6! 5!}} ... ?$$

Because, $6$ white balls and $5$ black balls are of same colors.

If my notion is incorrect, can you modify the same problem to show me in which case would it be applicable? i.e. $\binom {11}{3}$ would be divided by $6!5!$?

• You should include to your question why you think that it's true,so you can have precise answers. – Nameless Nov 2 '15 at 20:38
• Be specific about the title, that will facilitate finding this post to other users while searching. – user249332 Nov 2 '15 at 20:39
• $3$ balls are drawn from $6+5=11$ balls, so no. of way is $11\choose 3$, here, the orders of black and white balls are not important. Because order would make impact if there are $6$ white and $5$ black balls chosen, but here only $3$ balls are chosen, and you don't know the number Black and White balls in them, so, you need not divide them with $6!5!$. When, you will be told that how many ways the balls will be permuted, then the answer would be $\frac {11!}{6!5!}$. Did you understand? – user249332 Nov 2 '15 at 20:54
• @SubhadeepDey, But, combinations also use division by the number of common items. When? – user6704 Nov 2 '15 at 20:57
• @anonymous,Suppose that you are given the problem: In a regular $n$-gon ,how many diagonals are there? The approach should be following, From each vertices, you can draw exactly $n-3$ diagonals, and there are $n$ vertices. So, the no. of diagonals should be $n(n-3)$, but it's not. Note that, all the vertices are counted twice in that method. So simply divide by $2$ . hence the answer is $\frac {n(n-3)}2$. Here the division principle occurs. – user249332 Nov 2 '15 at 21:06

If you consider the indistinction of same-coloured balls in the denominator, you must also do so in the numerator.   ("As above, also below.")

To consider the indistinction of same-coloured balls, we look instead at ways to select places for white balls out of a line up of eleven, then ask what is the probability that the first three in any such line-up are two black and one white.

There are $\frac{11!}{6!5!}$ distinct ways to select six of eleven places for the white balls.

There are $\frac{3!}{1!2!}\frac{8!}{5!3!}$ distinct ways to select one of the first three and five of the remaining eight places for the white balls.

So the probability that only one white ball will be placed in the first three places is: $$\dfrac{\frac{3!}{2!1!}\frac{8!}{5!3!}}{\frac{11!}{6!5!}} = \frac{\frac{6!}{1!5!}\frac{5!}{2!3!}}{\frac{11!}{3!8!}}$$

So it turns out that is $\dfrac{\binom{6}{1}\binom{5}{2}}{\binom{11}{3}}$ after all.

• Somebody posted the same answer and then deleted it. – user6704 Nov 2 '15 at 21:31
• No, a somewhat different answer was posted, then retracted. It correctly noted that if indistinctness is accounted for in the denominator, then it also has to be so in the numerator. However, it could not show how this might be done properly. – Graham Kemp Nov 2 '15 at 21:43