Permutations (1 Viewer)

[goobi]

Question:

Find the total number of even numbers greater than 500 but less than 4000 which can be formed using the digits 1, 2, 3, 4, 5, 6 if NO digit can be repeated in any number.

Thanks for any help!

 

[deterministic]

Consider each case separately:
1) Three digit numbers - First digit must be a five or a six, and last digit must be 2,4,6 (even number). So deal with the restrictions first.
Suppose the first digit is 5, then there are 3 choices for the last digit (2,4,6), and hence for the middle digit, there will be 4 choices (2 even digits you didnt choose,1,3). So there are 3*4=12 possible 3 digit numbers starting with 5.
Suppose the first digit is 6, then there are 2 choices for last digit (2,4) and hence 4 choices for middle digit (ones you haven't used). So there are 2*4=8 possible 3 digit numbers starting with 6.

Thus there are 20 possible three digit even numbers

2) Four digit numbers. There are 2 main cases:
-First digit is odd (1,3) - There are 3 choices again for the last digit (2,4,6). The second digit will have 4 possibilities, and there are 3 more choices for the third digit. So together there are 2*4*3*3=72 numbers for this case. (note each number in the product represents the number of choices for each digit and you always deal with restricted cases first.
-First digit is even (2) - There are 2 choices for the last digit (4,6), then you can choose 4 possibilities for the second digit, and there are 3 more choices for third digit. So together there are 1*4*3*2=24 numbers for this case.

Thus the total number of even numbers between 4000 and 500 is 24+72+20=116

 

[goobi]

Consider each case separately:
1) Three digit numbers - First digit must be a five or a six, and last digit must be 2,4,6 (even number). So deal with the restrictions first.
Suppose the first digit is 5, then there are 3 choices for the last digit (2,4,6), and hence for the middle digit, there will be 4 choices (2 even digits you didnt choose,1,3). So there are 3*4=12 possible 3 digit numbers starting with 5.
Suppose the first digit is 6, then there are 2 choices for last digit (2,4) and hence 4 choices for middle digit (ones you haven't used). So there are 2*4=8 possible 3 digit numbers starting with 6.

Thus there are 20 possible three digit even numbers

2) Four digit numbers. There are 2 main cases:
-First digit is odd (1,3) - There are 3 choices again for the last digit (2,4,6). The second digit will have 4 possibilities, and there are 3 more choices for the third digit. So together there are 2*4*3*3=72 numbers for this case. (note each number in the product represents the number of choices for each digit and you always deal with restricted cases first.
-First digit is even (2) - There are 2 choices for the last digit (4,6), then you can choose 4 possibilities for the second digit, and there are 3 more choices for third digit. So together there are 1*4*3*2=24 numbers for this case.

Thus the total number of even numbers between 4000 and 500 is 24+72+20=116

Thank you so much for the excellent explanation