How many numbers are there between 100 and 1000 , in which all the digits are distinct?a)729b)648c)858d)546Correct answer is option 'B'. Can you explain this answer?

Question

Answer

H T U

9.9.8 = 9 x 9 x 8 = 648 numbers

No. of digits in hundreds place should be filled first as the zero should be excluded

(1,2,3,4,5,6,7,8,9) = 9 ways

No. of digits in tens place (now the zero can be included but one digit has been used at hundreds place) so, no. of ways we can fill ten's place is = 9 ways

No of digits in units place is any one among the remaining 8 no. = 8 ways

so, 9 x 9 x 8 = 648

Answer

Use permutation

Answer

We have to find the numbers between 100 and 1000 in which all digits are distinct
_ _ _ the first place can be filled in 9 ways i.e, 1to 9 ( if we place 0 on first place it will become two digited number)
the second place can be filled with 9 ways (as 0 can also be placed but...the number which is placed by on first place caned not be placed again)
the third place can be filled with 8 ways (as the numbers placed can not be placed again)
so
9×9×8=684

Answer

There are several steps we can follow to solve this problem:

1. Finding the range:
The problem states that we need to find the numbers between 100 and 1000. So, the range of numbers we need to consider is from 101 to 999. Since we want to find the numbers in which all the digits are distinct, we need to exclude any numbers that have repeated digits.

2. Counting the possibilities for each digit:
For the hundreds digit, we have 9 possibilities (1-9) since it cannot be zero.
For the tens and units digits, we have 9 possibilities for each (0-9), excluding the number already chosen for the hundreds digit.

3. Finding the total number of possibilities:
To find the total number of possibilities, we multiply the number of possibilities for each digit together. Therefore, the total number of possibilities is 9 * 9 * 8 = 648.

So, the correct answer is option 'B', 648.

Answer

3digit no....is...filling on  3 place... now first an b fll by 9...ways ...xluding 0....second place by 9 place as...0 can be fll in sec place...now ...v have only 8 digits left...hence ...last can b fll by 8 digits....so ttl ways = 9*9*8= 648

Similar JEE Doubts

How many numbers between 99 and 1000 can be formed from the digits 2,3,7,0,8,6 so that in each number each digit may occur once only?

How many numbers between 99 and 1000 can be formed from the digits 2,3,7,0,8,6 so that in each number each digit may occur once only?

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