A box contains two white marbles, three black marbles and four red marbles. In how many ways can three marbles be drawn from the box, if at least one black marble is to be included in the draw?
At least $1$ black marble is to be included in the draw - i.e. we can have $1$ OR $2$ OR $3$ black marbles in the draw.
Total number of marbles $= 2 + 3 + 4 = 9$
When $1$ black marble is selected
$1$ black marble can be selected from $3$ black marbles in $3$ ways
Then, $2$ marbles can be selected from the remaining $6$ ($2$ white and $4$ red) marbles in
${^{6}}{C}_{2} = \dfrac{6!}{4! \times 2!} = \dfrac{6 \times 5}{2} = 15$ ways
$\therefore \;$ Number of ways of selecting $3$ marbles when $1$ black marble is selected $= 3 \times 15 = 45$ ways
When $2$ black marbles are selected
$2$ black marbles can be selected from $3$ black marbles in ${^{3}}{C}_{2} = 3$ ways
Then, $1$ marble can be selected from the remaining $6$ ($2$ white and $4$ red) marbles in
${^{6}}{C}_{1} = 6$ ways
$\therefore \;$ Number of ways of selecting $3$ marbles when $2$ black marbles are selected $= 3 \times 6 = 18$ ways
When $3$ black marbles are selected
$3$ black marbles can be selected from $3$ black marbles in $1$ way
$\therefore \;$ Number of ways of selecting $3$ marbles so that at least $1$ black marble is selected
$= 45 + 18 + 1 = 64$ ways