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How many numbers between 5000 and 10000 can be formed using the digits 1, 2, 3, 4, 5, 6, 7, 8, 9 each digit appearing not more than once in each number?

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$5 \times^{8} P_{3}$

$5 \times^{8} C_{8}$

$5! \times^{8} P_{3}$

$5 \times^{8} C_{3} \times 3!$

answer is A, D.

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Detailed Solution

We are to form 4-digit numbers with either 5 or 6 or 7 or 8 or 9 at the thousand's place.

4-digit number : $\begin{array}{llll} X X X X \\ ↓ ↓ ↓ ↓ \\ 1 2 3 4 \end{array}$

No. of ways of filling 1st place = 5 After filling 1st -place, we have 8 digits to be used.

No. of ways of filling 2 nd , 3 rd, 4 th places = $^{8} P_{3}$

$∴$ Total no. of numbers formed $= 5 \times^{8} P_{3}$

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