How many 3 digit even numbers can be formed from?

60565248

Solution : No. of digits to be filled at one's place = 3
No. of digits to be filled at 10's place = 5
No. of digits to be filled at 100's place = 4
`:.` Total no. of digits formed `= 3 xx 5 xx 4 = 60`
If zero is at 100's place,
Then, no. of digits to be filled at one's place = 2
& no. of digits to be filled at 10's place = 4
`:.` No. of digits formed with zero at 100's place
`= 1 xx 2 xx 4 = 8`
`:.` Required no. of digits formed `= 60 - 8 = 52`

Solution : total number of digits = 6
The unit place can be filled with any one of the digits 2, 4, 6.
So number of permutation = `3P_1` = 3
Now, the tens and hundreds place can be filled by remaining 5 digits.
So number of permutations =`5P_2`= 20
Hence total number of permutations
= 3×20
=60.

Ex 7.3, 3 (Method 1) How many 3-digit even numbers can be made using the digits 1, 2, 3, 4, 6, 7, if no digit is repeated? We need to find 3 digit even number using 1, 2, 3, 4, 6, 7, Hence units place can have either 2, 4 or 6 Number of even numbers if 2 is at units place Hence these are 5 more digits left (1, 3, 4, 6, 7) for Hence n = 5 which we need to fill 2 place and r = 2 Number of 3 digit even number with 2 at unit place = nPr = 5P2 = 5!/((5 − 2)!) = 5!/3! = (5 × 4 × 3!)/3! = 20 Thus, Number of 3 digit even number with 2 at unit place = 20 Similarly Number of 3 digit can number with 4 at unit place = 20 and 6 at unit place = 20 Hence, Total 3-digit even numbers = 20 + 20 + 20 = 60 Ex 7.3, 3 (Method 2) How many 3-digit even numbers can be made using the digits 1, 2, 3, 4, 6, 7, if no digit is repeated? Let the 3 digit even number be Only 3 numbers are possible at units place (2 , 4 & 6) as we need even number. Number of 3 digit even numbers = 3 × 5 × 4 = 60

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