Question

Find the number of integers greater than $4000$ that can be formed by using the digits $3,4,5$ and $2$ if every digit can occur at most once in any number

• A. $6$
• B. $18$
• C. $12$
• D. $24$

Answer 1

The correct option is C $12$

Since each desired number is to be greater than $4000$, so we must have $4or5$ at the thousands place.
Case 1, when $4$ is at the thousands place
Then the hundreds, tens and ones place can be filled up by remaining $3$ digits in ${}^3{P}_3=\frac{3!}{(3-3)!}=\frac{3!}{0!}=3\times 2=6$ ways
Case 2, when $5$ is at the thousands place
Then the hundreds, tens and ones place can be filled up by remaining $3$ digits in ${}^3{P}_3=\frac{3!}{(3-3)!}=\frac{3!}{0!}=3\times 2=6$ ways
So, total number ways $=6+6=12$ ways