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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.

9P3=9!(9−3)!=9!6!

Number of such 3-digit numbers

9×8×7×6!6!=9×8×7=504

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

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