How-many-numbers-from-1-to-999-do-n ot-have-any-repeated-digits? And how-many-are-even?

• The single-digit numbers 1-9 do not have any repeated digits, so they contribute 9 numbers.

• Of double-digit numbers 10-99, the first digit can be picked from 1-9 and the second from 0-9 minus whichever one is picked for the first digit, so they contribute 9×(10 - 1) = 81 numbers.

• Of the triple-digit numbers 100-999, the first digit is picked from 1-9 and the second and third digits from 0-9, so they contribute 9×(10 - 1)×(10 - 2) = 648 numbers.

So there are 9 + 81 + 648 = 738 numbers from 1-999 without repeated digits.

Of these numbers, the ones that are even end in 0, 2, 4, 6, or 8.

• Of the single-digits, there are 4 numbers.

• Of the double-digits, the first digit can be chosen from 1-9 and the second from {0, 2, 4, 6, 8}.

•• If the first digit is odd and picked from {1, 3, 5, 7, 9}, then they contribute 5×5 = 25 numbers.

•• If the first digit is even, then they contribute 4×(5 - 1) = 16 numbers (because the first digit can't be 0).

• Of the triple-digits, the third digit is picked from {0, 2, 4, 6, 8}.

•• If both the first and second digits are odd, then they contribute 5×(5 - 1)×5 = 100 numbers.

•• If the first is odd and the second is even, then they contribute 5×5×(5 - 1) = 100 numbers.

•• If the first is even and the second is odd, then they contribute 4×5×(5 - 1) = 80 numbers.

•• If both are even, then they contribute 4×(5 - 1)×(5 - 2) = 48 numbers.

So there are 4 + (25 + 16) + (100 + 100 + 80 + 48) = 373 even numbers from 1-999 with no repeating digits.

answer: 34rtyj,jkhtyetwrhyjeuiejnbhdfknbkguxfrbhkfdbjvkmbd,kvbsd.kxjvbhjnfbgjxdfn

step-by-step explanation:

that is the sum

step-by-step explanation:

there are no other answers without knowing the variables values