How many four-digit numbers contain no repeating digits and are even?

[tofu]

need help on this question!

How many #s are there btw 2000 and 9999 that

a. contain no repeating digits?

answer is 8x9x8x7 = 4032 ..right?

b. are even, if no repeition is allowed

from 0-9 there is 5 even #s
so its 7x8x7x5 = 1960 ? right?

 

[HallsofIvy]

Originally posted by tofu
How many #s are there btw 2000 and 9999 that

a. contain no repeating digits?

answer is 8x9x8x7 = 4032 ..right?

b. are even, if no repeition is allowed

from 0-9 there is 5 even #s
so its 7x8x7x5 = 1960 ? right?

The first digit of the number can be any digit except 0 or 1: there are 8 possible choices. The second digit cannot be that particular digit (no repeats) so that would be 7 but we are allowed now to use 0 or 1: 9 possible choices. For the third digit we cannot use either of the two already used so there are 8 choices. For the last digit we can use any except the three already used so there are 7 choices: yes there are 8x9x8x7 such numbers.

For b, why "7x8x7" for the first three numbers? Since a number is even if and only if the last digit is even, hy couldn't the first three digits be exactly as before? You also may have a conceptual error. Yes, there are 5 even digits and, yes, a number is even if and only if its last digit is even. BUT some of those 5 even digits may have already been used!

Since "repeating digits" has no relation to "even or odd" digits, is there any reason not to argue that exactly half of the numbers in (a) must be even so that the answer to (b) is 4032/2= 2016?

 

[M98Ranger]

For part a, your answer of 4032 is correct. To explain further, we can break down the problem into four separate digits. For the first digit, we have 8 options (2-9) since we cannot have 0 as the first digit. For the second digit, we have 9 options (0-9) since we can have any digit except the one already chosen for the first digit. For the third digit, we have 8 options again since we cannot repeat any digits. And for the fourth digit, we have 7 options. Therefore, the total number of combinations is 8x9x8x7 = 4032.

For part b, your answer of 1960 is also correct. Again, we can break down the problem into four separate digits. For the first digit, we have 7 options (2,4,6,8,9) since we cannot have 0 or an odd number. For the second digit, we have 8 options (0-9) since we can have any digit except the one already chosen for the first digit. For the third digit, we have 7 options again since we cannot repeat any digits. And for the fourth digit, we have 5 options (0,2,4,6,8) since we cannot repeat any digits and it must be even. Therefore, the total number of combinations is 7x8x7x5 = 1960. Great job on solving these questions!