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The number of all numbers having 5 digits, with distinct digits is
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Verified Answer
The correct answer is:
$9 \times{ }^{9} P_{4}$
Total number of 5-digit numbers having all the digits distinct $={ }^{10} \mathrm{P}_{5}-{ }^{9} \mathrm{P}_{4}$
$$
\begin{array}{l}
=\frac{10 !}{5 !}-\frac{91}{5 !}=\frac{10 \times 9 !}{5 !}-\frac{91}{51} \\
=\frac{91}{51}(10-1)=9 \times \frac{9!}{(9-4) !} \\
=9 \times 9 P_{4}
\end{array}
$$
$$
\begin{array}{l}
=\frac{10 !}{5 !}-\frac{91}{5 !}=\frac{10 \times 9 !}{5 !}-\frac{91}{51} \\
=\frac{91}{51}(10-1)=9 \times \frac{9!}{(9-4) !} \\
=9 \times 9 P_{4}
\end{array}
$$
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