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
Show More10*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`.