Question

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

• A. 96
• B. 120
• C. 312
• D. 360

Answer 1

The correct option is C

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

$\underset{5}{\stackrel{-}{|}}\underset{4}{\stackrel{-}{|}}\underset{3}{\stackrel{-}{|}}\underset{2}{\stackrel{\_0}{|}}$

So, the number of possible numbers = $5\times 4\times 3\times 2$ = 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.

$\underset{4}{\stackrel{-}{|}}\underset{4}{\stackrel{-}{|}}\underset{3}{\stackrel{-}{|}}\underset{2}{\stackrel{\_2}{|}}$

So, the number of possible numbers = $4\times 4\times 3\times 2$ = 96

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

$\underset{4}{\stackrel{-}{|}}\underset{4}{\stackrel{-}{|}}\underset{3}{\stackrel{-}{|}}\underset{2}{\stackrel{\_4}{|}}$

Number of possible numbers = $4\times 4\times 3\times 2$ = 96

$\therefore$ the total number of arrangements = 120 + 96 + 96 = 312