Question

The number of integers greater than 2000 that can be formed with the digits 0,1,2,3,4,5 using each digit at most once,is

• A. 1440
• B. 1200
• C. 1420
• D. 1480

Answer 1

Answer: (A)

1440

Answer 2

The correct answer is A:1440
We are given the digits 0,1,2,3,4, and 5,and we need to form integers greater than 2000 using these digits, with each digit being used at most once.
Case 1: Integers between 2000 and 2999
For the thousands place, we can choose any of the 6 digits except 0(0 cannot be the first digit).So, there are 5 choices.
For the hundreds place, we have 5 remaining digits to choose from (since one digit has already been used in the thousands place).
For the tens place, we have 4 remaining digits to choose from.
For the units place, we have 3 remaining digits to choose from.
Total integers in this range=$(5 \times 5 \times 4 \times 3 = 300) integers$.
Case 2 : Integers between 3000 and 4999
For the thousands place, we can choose any of the 4 digits other than 0(since 0 cannot be the first digit anymore).
For the hundreds place, we have 5 remaining digits to choose from.
For the tens place, we have 4 remaining digits to choose from.
For the units place, we have 3 remaining digits to choose from.
Total integers in this range=$(4 \times 5 \times 4 \times 3 = 240) integers$.
Case 3: Integers between 5000 and 5999
For the thousands place, we can choose any of the 2 digits other than 0.
For the hundreds place, we have 5 remaining digits to choose from.
For the tens place, we have 4 remaining digits to choose from.
For the units place, we have 3 remaining digits to choose from.
Total integers in this range=$(2 \times 5 \times 4 \times 3 = 120) integers$.
Total: Adding up the integers from all three ranges:(300+240+120=660) integers.

However, we need to consider that there are 6 different digits available, and we can arrange them in $(6!)$ ways.
This includes repetitions, which we need to exclude. So, the final answer is $(6!-660=1440)$ integers.
Hence, the correct answer is 1440.