NIMCET 2014 — Mathematics PYQ

To find the total number of even integers with distinct digits, we must consider the constraints on the thousand's place, the hundred's place, the ten's place, and the unit's digit.

Calculation

In the range $4000$ to $7000$, the thousand's place digit can be $4$, $5$, or $6$. We must split this into two cases based on whether the thousand's digit is even or odd.

Case 1: Thousand's place digit is even (i.e., $4$ or $6$)

• Thousand's place: There are $2$ choices ($4$ or $6$).

• Unit's place: Since the number must be even, the unit's digit must be from $\{0,2,4,6,8\}$. However, one even digit is already used in the thousand's place, leaving only $4$ possibilities for the unit's digit.

• Hundred's place: There are $8$ remaining digits available (out of $10$, after using two).

• Ten's place: There are $7$ remaining digits available.

• Total for Case 1: $2\times 8\times 7\times 4=448$.

Case 2: Thousand's place digit is odd (i.e., $5$)

• Thousand's place: There is only $1$ choice ($5$).

• Unit's place: There are $5$ choices for the unit's digit to make it even ($\{0,2,4,6,8\}$).

• Hundred's place: There are $8$ remaining digits available.

• Ten's place: There are $7$ remaining digits available.

• Total for Case 2: $1\times 8\times 7\times 5=280$ possibilities.

Total number of even integers:

Correct Option: 4