Q26 / 48 · JEE Advanced 2025 (Paper 1) · Mathematics
Let $S$ be the set of all seven-digit numbers that can be formed using the digits 0,1 and 2 . For example, 2210222 is in $S$, but 0210222 is NOT in $S$.
Then the number of elements $x$ in $S$ such that at least one the digits 0 and 1 appears exactly twice in $x$, is equal to ______.
Correct answer: 762
Solution
Let $\mathrm{A} \rightarrow$ " 0 " appear exactly twice.
and $\mathrm{B} \rightarrow " 1 "$ appear exactly twice.
$\therefore \mathrm{A} \cap \mathrm{B} \rightarrow$ " 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
$\underline{1}\underbrace{-------}$
$\text { Number of ways }=\underset{\text { placing } 1}{} \mathrm{}^6{C}_1(1)(2)^5=192$
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 $\mathrm{n}(\mathrm{A} \cap \mathrm{B})=\underset{\text { placing zero }}{{ }^6 \mathrm{C}_2(1)} \times \underset{\text { placing } 1}{{ }^5 \mathrm{C}_2(1)} \times \underset{\text { 2at rest places } 1}{(1)} =\frac{6 \times 5}{2} \times \frac{5 \times 4}{2}=150$
$\begin{aligned} & \therefore \mathrm{n}(\mathrm{A} \cup \mathrm{B})=\mathrm{n}(\mathrm{A})+\mathrm{n}(\mathrm{B})-\mathrm{n}(\mathrm{A} \cap \mathrm{B}) \\ & =480+(192+240)-150 \\ & =762\end{aligned}$
Hence, the correct answer is 762.