How many 5-digit even numbers without repetition of digits can be formed from the digits: 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9?
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Lets divide the possibilities in two parts:-
Other part which is a collection of numbers which do not end with 0.
One part which is a collection of numbers which end with 0.
Part 1:-
If the number does not end with 0, the units digit can be filled in with the other even numbers which is in 4 ways. The ten thousandth digit can be filled in 8 ways becayse we cant place 0 and one digit has already been used up. Then the thousandth digit can be filled in 8 ways because 2 numbers has been used up and 0 can be used. The hundredth digit can be filled in 7 ways because 3 digits have already been used up. Lastly, the tenth digit can be filled in 6 ways because 4 digits have been used up. So, the number of even 5 digit numbers without repitition of digits which do not end with 0 that can be formed = 8 × 8 × 7 × 6 × 4 = 1 0 7 5 2 .
Part 2:-
If the number ends with 0 then we can fill the units digit in only 1 way. Then we can fill the ten thousandth digit in 9 ways. Then the thousandth in 8 ways, the hundredth in 7 ways and the tenth in 6 ways. So, the number of 5-digit even numbers without repitition of digits which end with 0 can be formed in 9 × 8 × 7 × 6 × 1 = 3 0 2 4 .
Since we can make numbers which either fall in part 1 or part 2, we add the results. So, the final answer is 1 0 7 5 2 + 3 0 2 4 = 1 3 7 7 6 .
Alternatively:-
Total number of 5 - digit odd numbers without repitition of digits can end in 5 ways. The ten thousandth digit can be filled in 8 ways because 0 cant be used and 1 number is already used up. So, the thousandth digit can be filled in 8 ways because 2 numbers are used up and 0 can be used now. The hundredth digit can be filled in 7 ways and the tenth digit can be filled in 6 ways which give the result as 8 × 8 × 7 × 6 × 5 = 1 3 4 4 0 .
Total 5-digit numbers without repitition of digits that can have their ten thousandth digit in 9 ways because 0 cant be used. Then the thousandth digit can be filled in 9 ways again because 1 number is used up and 0 can be used. Similarly, the hundredth digit can be filled in 8 ways and the tenth digit can be filled in 7 ways and the units digit in 6 ways which gives the result as 9 × 9 × 8 × 7 × 6 = 2 7 2 1 6 .
Now 5-digit numbers whose digits do not repeat are either even or odd. So, the 5-digit even numbers whose digits do not repeat can be obtained by subtracting the number of 5-digit odd numbers whose digits do not repeat from the total number of 5-digit numbera whose digits do not repeat. So, the answer is 2 7 2 1 6 − 1 3 4 4 0 = 1 3 7 7 6 .