Question

What are the total number of ways in which a
four digit number that is divisible by 2, can be
formed using the numerals 0, 1, 2, 3, and 4
without repetition?​

Answer 1

Answer:

4

Step-by-step explanation:

Case1: Take 0 at the units place

Remaining 3 places can be filled from the remaining 4 digits in 4P3 = 4!/(4–3)!=4!/1! = 24

Case2: Take 2 at the units place

As we require a 4 digit number, so the first digit cannot be chosen as 0 (zero). So the first digit can be 1, 3 or 4 i.e. 3 choices for the first place and then second place also 3 choices (2 from above and a 0) and finally 2 choices for the third place. This gives us, in all 3*3*2=18 such numbers.

Case3: Take 4 at the units place

This case is similar to the above case2. So in this case also there would be 18 numbers.

Hence in all there are 24+18+18=60 numbers satisfying above criteria.

Answer 2

The total numbers of ways is 60.

GIVEN

Four digit number that is divisible by 2, can be

formed using the numerals 0, 1, 2, 3, and 4 without repetition.

TO FIND

Total number of ways.

SOLUTION

We can simply solve the above problem as follows;

It is given that the 4-digit number formed should be divisible by 2. This means the digit in unit's place can be 0,2 or 4.

If zero is in the unit's place;

Combination of remaining 3 digits place = ⁴C₃ = 4!/(4-3)!

= 4!/1!

= 24.

If 2 is at unit's place.

Zero cannot be at thousandth's place.

The first digit can be 1, 3 or 4.

Therefore there are 3 choices for first digit.

3 choices for second place.

And,

2 choices for third place.

The combinations are- 3×3×2 = 18 combinations.

4 is at unit's place.

Combinations of numbers will be = 18.

Total combinations = 24 + 18 + 18 = 60 Numbers.

Hence, The total numbers of ways is 60.

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