Question

A box contains two white balls, three black balls and four red balls. In how many ways can three balls can be drawn from the box if atleast one black ball is to be included in the draw?

Answer 1

The possibilities of choosing atleast one black ball are

$1$black$+2$non-black.

or $2$black$+1$non-black

or $3$black$+0$non-black.

Number of ways of choosing $3$ balls with atleast one black ball

$={3_C}_1\times {6_c}_2+{3_c}_2\times {6_c}_1+{3_c}_3\times {6_c}_0$

We have $c(n,r)=\frac{n!}{r!(n-r)!}$

We have ${3_C}_1=\frac{3!}{1!2!}=\frac{3\times 2\times 1}{1\times 2\times 1}=3$

${3_C}_2=\frac{3!}{1!2!}=\frac{3\times 2\times 1}{1\times 2\times 1}=3$

${6_c}_2=\frac{6!}{2!4!}=\frac{6\times 5\times 4!}{2!4!}=\frac{6\times 5}{2}=15$

${6_c}_1=\frac{6!}{1!5!}=6$

${3_c}_3=\frac{3!}{3!\times 0!}=1$

Number of ways of choosing $3$ balls with atleast one black ball$=3\times 15+3\times 6+1$

$=45+18+1=64$