How many 3 digit numbers can be formed using 1-9

Ex 7.3, 1 (Method 1) How many 3-digit numbers can be formed by using the digits 1 to 9 if no digit is repeated? Let the 3-digit number be Number of 3 digit number which can be formed = 9 × 8 × 7 = 504 ways Ex 7.3, 1 (Method 2) How many 3-digit numbers can be formed by using the digits 1 to 9 if no digit is repeated? Given n = numbers from 1 to 9 = 9 r = 3 Required 3 digit numbers = 9P3 = 9!/(9 − 3)! = 9!/6! = (9 × 8 × 7 × 6!)/6! = 9 × 8 × 7 = 504

10*10*109*9*99*8*7None of these

Solution : 3 digit numbers that can be formed using the digits from 1 to 9.
Required number of 3-digit numbers
= arranging 3 digits with the total number of 9 digits.
= `9P_3` = 9×8×7.
C is the correct answer.

Solution : Given `3` digit numbers that can be formed using the digits from 1 to `9`.
Required number of `3`-digit numbers= arranging `3` digits with the total number of `9 ` digits..
=`9_(P_3)=(9xx8xx7)=504`..
Hence, the required number of numbers =`504`.

How many 3 digit numbers can be formed from 1 to 9 if it should be even without repetition?

There are 504 different 3-digit numbers which can be formed from numbers 1, 2, 3, 4, 5, 6, 7, 8, 9 if no repetition is allowed.

How many three digit numbers can be formed using the digits 0 to 9?

We have to find the number of arrangements of 3 digits which can be formed from digits 0 to 9. Therefore, in 1000 arrangements 3 digits can be formed from the digits 0 through 9.

3 digits