Question

how many 4-digit numbers can be formed from the digits 2, 3, 5, 0, 6, 8, 7 and 9, which are divisible by 2 and none of the digits is repeated?

Answer 1

Answer:

6,912

Step-by-step explanation:

Last Digit 2,0,6,8 are divisible by 2

and there are 4! ways..

Remaining 3 places are filled by remaining( 4!*3!*2!)

so final answer will 4!*(4!*3!*2!)

Answer 2

Answer:

The total number of ways a 4-digit number can be formed is 840.  

Step-by-step explanation:

Given the digits 2, 3, 5, 0, 6, 8, 7 and 9

Need to find 4-digit numbers that are divisible by 2 without repeating the digits.

The number of numbers given is 8.

In that a number to be divisible by 2, the last digit should be 0, 2,6 or 8.

So any one of the four can be there in the unit's place. Therefore there are four ways of chosing the number.

Once the unit's place is chosen, then there are 3 places left, which can be taken by the remaining 7  numbers.

                            $\therefore\ \ ^{7} P_{3} =\frac{7!}{(7-3)!}=\frac{7!}{4!}$

                                         $=7\times 6 \times 5=210$

The total number of ways a 4-digit number can be formed is

                                   $4\times 210=840$

Hence, the total number of ways a 4-digit number can be formed is 840.