Question

How many 4-digit numbers are possible with the digits 0 to 9 and also it should be even no​

Answer 1

Answer:

For the first digit place there are 9 options , second digit place there are 8 options, third digit place there are 7 options and fourth digit place there are 6 options.

so number of 4 digit numbers=9×8×7×6=3024

Step-by-step explanation:

Answer 2

Step-by-step explanation:

(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 5-digits.(because we can say any number even if their once place ha 0,2,4,6,8)

Thus ,the number of possibilities =5

Hence the total number of possibilities to arrange the even numbers from 0 to 9 without repetition of any digits =9X9X8X5=3240ways.

case2:if repetition is allowed then

9×10×10×5=4500 ways

hope you can understand.

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