How many 5 digit even numbers can be formed from digits 0 1,2 3 and 4 if repetition of digit is not allowed?

Using the digits 1, 2, 3, 5, and 6, without repetition, how many five-digit even numbers can be formed?

1. 51
2. 24
3. 48
4. 96
5. 64

Concept:

Basic Principle of Counting:

If there are m ways for happening of an event A, and corresponding to each possibility there are n ways for happening of event B, then the total number of different possibilities for happening of events A and B are:

• Either event A OR event B alone = m + n.
• Both event A AND event B together = m × n.
• An even number's units digit is either 0, 2, 4, 6 or 8.

Calculation:

For an even number, the units place must be 2 or 6, so it can be filled in 2 ways.

The remaining four places can now be filled in 4, 3, 2 and 1 ways respectively.

The total number of ways in which the number can be written = 4 × 3 × 2 × 1 × 2 = 48.

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Solution : Total given digits = 5
We have to place these 5 digits at unit, tens, 100th, 1000th and 10000th place.
We cannot take 'O' at 1000th place, so this place can be filled in 4 ways.
Repetition of digits is allowed.
`:.` Each place out of unit, 10th, 100th and 1000th can be filled in 5 ways.
Now form multiplication rule
Total numbers `= 4 xx 5 xx 5 xx 5 xx 5 = 2500`.

The correct option is D

312

Even numbers can be formed by placing either 0 or 2 or 4 in the rightmost digit.

_ _ _ _ 0

_ _ _ _ 2

_ _ _ _ 4

So, let us consider all the 3 cases separately.

Case 1: Rightmost digit is 0
Available digits - 1,2,3,4,5

Leftmost digit can be filed with 1 or 2 or 3 or 4 or 5

So, number if ways of filling leftmost digit = 5

As 2 digits are already used, the next position can be filled with either of the 4 digits.

Proceeding in this manner, the next 2 digits can be filled in 3 and 2 ways respectively

−|5−|4−|3_0|2

So, the number of possible numbers = 5×4×3×2 = 120

Case 2: Rightmost digit is 2

_ _ _ _ 2 Available digits 0,1,3,4,5

Out of these 5 digits leftmost digit can be filed with 1 or 3 or 4 or 5
As if 0 comes on the left most position the number will become a 4 digit number.
So, number if ways of filling leftmost digit = 4
As 2 digits are already used, the next position can be filled with either of the 4 digits .

Proceeding in this manner, the next 2 digits can be filled in 3 and 2 ways respectively.

−|4−|4−|3_2|2

So, the number of possible numbers = 4×4×3×2 = 96

Case 3: Rightmost digit is 4
This case is similar to 2nd case.

−|4−|4−|3_4|2

Number of possible numbers = 4×4×3×2 = 96

∴ the total number of arrangements = 120 + 96 + 96 = 312

How many 5 digit even numbers can be formed using the digits 1,2 3 4 5 exactly once?

How many 5 digit even numbers can be formed from the digits 0 1,2 3 and 4 if repetition of digit is not allowed?

How many five

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