Question

Open in App

Solution

Verified by Toppr

$□□□□T_{h}HTO$

Since, leftmost place i.e thousand's place cannot have zero. So, there are $9$ ways to fill thousand's place.

Since, repetition is not allowed , so hundreds place can be filled by remaining $9$ digits .

So, hundred's place can be filled in $9$ ways.

Similarly, to fill tens place, we have $8$ digits remaining.

So,tens place can be filled by $8$ ways.

Similarly, to fill ones place, we have $7$ digits remaining.

So,ones place can be filled by $7$ ways.

So, required number of ways in which four digit numbers can be formed from the given digits is $9×9×8×7=4536$

The thousands place of the $4$-digit number is to be filled with any of the digits from $1$ to $9$ as the digit $0$ cannot be included. Therefore, the number of ways in which thousands place can be filled is $9$.

The hundreds, tens, and units place can be filled by any of the digits from $0$ to $9$.

However, the digits cannot be repeated In the $4$-digit numbers and thousands place is already occupied with a digit. The hundreds, tens, and units place is to be filled by the remaining $9$ digits.

Therefore, there will be as many such $3$-digit numbers as there are permutations of $9$ different digits taken $3$ at a time.

$_{9}P_{3}=(9−3)!9! =6!9! $

Number of such $3$-digit numbers

$6!9×8×7×6! =9×8×7=504$

Thus, by multiplication principle, the required number of $4$-digit numbers is $9×504=4536$

Video Explanation

Was this answer helpful?

0

0