AP EAMCET · Maths · Permutation Combination
The number of numbers lying between 1000 and 10000 such that every number contains the digits 3 and 7 only once without repetition is
- A 1140
- B 918
- C 720
- D 810
Answer & Solution
Correct Answer
(C) 720
Step-by-step Solution
Detailed explanation
No. of ways \(\underline{0} \ldots={ }^3 \mathrm{C}_2 \times 2 \times 8=48\) Total no. of ways \(\begin{aligned} & ={ }^4 C_2 \times 2 \times 8 \times 8=64 \times 12=768 \\ & \text { Required number }=768-48=720 \end{aligned}\)
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