Mathematics - Permutation & Combination Question with Solution | TestHub

Atleast one black ball can be drawn in the following ways (1) one black and two other colour balls $$={}^3{\mathrm{C}}_1\times {}^6{\mathrm{C}}_2=3\times 15=45$$ (ii) two black and one other colour balls $$={}^3{\mathrm{C}}_2\times {}^6{\mathrm{C}}_1=3\times 6=18$$ (iii) All the three are black $={}^3{\mathrm{C}}_3\times {}^6{\mathrm{C}}_0=1$ $\therefore$ Req. no. of ways $=45+18+1=64$