The number of numbers lying between 1000 and 10000 such that every number contains the digits 3 and 7 only…
The number of numbers lying between 1000 and 10000 such that every number contains the digits 3 and 7 only once without repetition is
- 1140
- 918
- 720
- 810
Solution
No. of ways $\underline{0} \ldots={ }^3 \mathrm{C}_2 \times 2 \times 8=48$
Total no. of ways
$\begin{aligned}
& ={ }^4 C_2 \times 2 \times 8 \times 8=64 \times 12=768 \\
& \text { Required number }=768-48=720
\end{aligned}$
Asked in: AP EAMCET 2024 (20 May Shift 2)
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