# Difficult Permutation-Combination Solved QuestionAptitude Discussion

 Q. 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. 499 ✖ B. 500 ✖ C. 375 ✔ D. 376 ✖ E. 501

Solution:
Option(D) is correct

The smallest number in the series is 1000, a 4-digit number.

The largest number in the series is 4000, the only 4-digit number to start with 4.

The left most digit (thousands place) of each of the 4 digit numbers other than 4000 can take one of the 3 values 1 or 2 or 3.

The next 3 digits (hundreds, tens and units place) can take any of the 5 values 0 or 1 or 2 or 3 or 4.

Hence, there are $3 × 5 × 5 × 5$ or 375 numbers from 1000 to 3999.

Including 4000, there will be $\textbf{376}$ such numbers.

## (2) Comment(s)

Anmol Seth
()

According to me it answer must be 375

As we fix 1 on first position and than $5 \times 5 \times 5 =125$ than fix 2 and 3 than answer will be $125 \times 3= 375$

Genius Genius
()

You a innocent boy, you forgetting the number 4000 can be an answer.