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Bunuel
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999< x <=4000

(1,2,3) (0,1,2,3,4) (0,1,2,3,4) (0,1,2,3,4)
—— —— —— ——
(3)*(5)*(5)*(5)

Total possibilities =3*5*5*5+1 =376
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a number must be greater than 999 and less than 4000
Digits we have 0,1,2,3,4 (repeat allowed )
So for the first we have 3 options(1,2,3) for rest of the 3 position we have 5 options each.

3*5*5*5=375
The one number we did not consider is number starting with 4 but only such no. is possible (4000)

so the number of such integers should be 375+1=376 (option B)
Bunuel
How many integers, greater than 999 but not greater than 4000, can be formed with the digits 0, 1, 2, 3 and 4, if repetition of digits is allowed?

A. 375
B. 376
C. 499
D. 500
E. 501

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