Mathematics - Permutation & Combination Question with Solution | TestHub

Let $\mathrm{A}\to$ " 0 " appear exactly twice. and $\mathrm{B}\to "1"$ appear exactly twice. $\therefore \mathrm{A}\cap \mathrm{B}\to$ " 0 " and " 1 " both appears exactly twice. $\mathrm{n}(\mathrm{A})=-\underbrace{-------}$ $={}_{\text{placing zero }}^6{\mathrm{C}}_2(1)(2{)}^5=\frac{6\times 5}{2}\times 2^5=480$ for $\mathrm{n}(\mathrm{B})$ C-I : $1$ at first place $\underline{1}\underbrace{-------}$ $\text{ Number of ways }=\underset{\text{ placing }1}{}{}^6{C}_1(1)(2{)}^5=192$ C-II : 2 at first place $\underline{2}\underbrace{-------}$ Number of ways $=\underset{\text{ placing }1}{{}^6{\mathrm{C}}_2(1)(2{)}^4}=\frac{6\times 5}{2}\times 2^4=240$ $n(B)=240+192$ for $\mathrm{n}(\mathrm{A}\cap \mathrm{B})$ $\underline{}\underbrace{-------}$ for $\mathrm{n}(\mathrm{A}\cap \mathrm{B})=\underset{\text{ placing zero }}{{}^6{\mathrm{C}}_2(1)}\times \underset{\text{ placing }1}{{}^5{\mathrm{C}}_2(1)}\times \underset{\text{ 2at rest places }1}{(1)}=\frac{6\times 5}{2}\times \frac{5\times 4}{2}=150$ $\begin{array}{rl}& \therefore \mathrm{n}(\mathrm{A}\cup \mathrm{B})=\mathrm{n}(\mathrm{A})+\mathrm{n}(\mathrm{B})-\mathrm{n}(\mathrm{A}\cap \mathrm{B})\\ & =480+(192+240)-150\\ & =762\end{array}$