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Shouldn't it should be mentioned that no digit can be repeated?
Or we need to assume no digit can be repeated unless given otherwise

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Edited to avoid confusion. Thank you.
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3×4×3×2=72
Option C

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There are 3 numbers that can take the first place - 7 8 and 9.
Consequently, 4 numbers can take the 100s place
3 can take the 10s place
and 2 can take the units place.

3*4P3 = 72


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First digit can only be 7,8 or 9 in order to be > 7000

so we get: 3C1 * 4C1 * 3C1 * 2C1 * 1 for all digits => results in 72 possibilities.
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Is the OA correct? Please post the OE

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This is a permutation problem.

How many 4-digit numbers greater than 7,000 can be formed from the digits 3, 5, 7, 8 and 9? (Assume no digit can be repeated)

A. 48
B. 60
C. 72
D. 80
E. 82

Since the numbers have to be greater than 7. The first digit can only be 7, 8, or 9.

7 _ _ _ <--- 4P3 = 4! / (4-3)! = 4 x 3 x 2 x 1 = 24 <--- Total # of ways we can get a digit greater than 7000 but less than 8000
8 _ _ _ <--- 4P3 = 4! / (4-3)! = 4 x 3 x 2 x 1 = 24 <--- Total # of ways we can get a digit greater than 8000 but less than 9000
9 _ _ _ <--- 4P3 = 4! / (4-3)! = 4 x 3 x 2 x 1 = 24 <--- Total # of ways we can get a digit greater than 9000 but less than 10000

24 + 24 + 24 = 72

Ans is C
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