Answer:

The Question can be re-written as :How many 4-digit numbers are possible with the digits 0 to 9?

(I)Digits cannot be repeated

Solution: There are 10-digits :0,1,2,3,4,5,6,7,8,9

The digits to be formed =No.of places=4

(I)Case I: Digits cannot be repeated:If 0 is placed in first place then it becomes a 3-digit number out of 4-places.Thus ,we can fill 1 or 2 or 3 or 4 or 5 or 6 or 7 or 8 or 9 in the first place.

Therefore,No.of possibilities in the first place =9

Again,consider the second place.Here we can fill 0 and any of the eight digits

Thus, No.of possibilities=9 (the digit 0 and 8 digits)

Consider the third place.We can fill any of the 8 digits.

Thus, No.of possibilities=8

Consider the fourth place.Here we can fill any 7-digits.

Thus ,the number of possibilities =7

Hence the total number of possibilities to arrange the even numbers from 0 to 9 without repetition of any digits =9X9X8X7=4536 ways.

Answer:

Solution

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Correct option is C)

For even no, last digit could

be 0,2,4,6

But first digit cannot be 0

So : I:6×5×4=120

II:5×5×4=100

III:5×5×4=100

IV:5×5×4=100

∴ Total no.of 4 digit even

no. =120+100+100+100

=420

Step-by-step explanation:

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