I have a question in my textbook.
A box contains $2$ white balls, $3$ black balls and $4$ red balls. In how many ways can $3$ balls be drawn from the box, if at least one black ball is to be included in the draw?
My assumption was that the ways to draw the fixed 'at least $1$ black ball' were same as the ways to draw $2$ or $3$ black balls, i.e. $1$. So, for each case;
$1$ black ball, and $2$ others, which is $1\times\binom62$
$2$ black ball, and $1$ other, which is $1\times\binom61$
$3$ black ball, and $0$ others, which is $1\times\binom60$
For a total of $22$ cases. But the solution in my textbook assumes otherwise. That each black ball is distinct. Is there something I missed in the statement or it is ambiguous and correct both ways?