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    				| 1. | 
                                    How many 3-digit even numbers can be made using the digits 1, 2, 3, 4, 6, 7 if no digit is repeated? | 
                            
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Answer»  Here total number of digits = 6 The unit place can be filled with any one of the digits 2, 4, 6. So number of permutation = 3P1 = 3!/2! = 3 Now the tens and hundred place can be filled by remaining 5 digits. So number of permutation = 5P3 = 5!/3! = (5 x 4 x 3!)/3! = 20 Hence total number of permutations = 3 x 20 = 60  | 
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