How many 3 digit even numbers can be formed from the digits 1,2,3,4,5,6
1.If the digits are not repeated
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\[n!/ (n-r)!r!\]\[\rightarrow 6! /(6-3)!3! =20\]
For the third spot, you only have 3 choices (2,4,6). For the rest of the two spots you have 5 choices. 5P2*3? maybe
1st, think of the last digit, as it is an even no. there are only 3 choices can fulfill the requirement
Second, think of the 1st digit, it can be any number except the last digit, which has been chosen already, so you have 5 choices
3rd, think of the 2nd digit, it can be any number except the last digit and the 1st digit, which have been chosen already, so you have 4 choices
finally , multiply the choices of each case together. that is 3x5x4 = 60 numbers
not sure if it is correct