How many $3$-digit even numbers less than $700$ can be formed from the digits $0, 2, 4, 5, 6, 8$ if repetition of digits is not allowed?

Question

Consider the digits $0,2,4,5,6,8$. How many $3$-digit even numbers less than $700$ can be formed if repetition of digits is not allowed? Note the first digit can't be zero.

Answer 1

Let the set of $3$ digit even numbers be represented by _ _ _.

Last digit can be $0, 2, 4, 6 \text{ or } 8$; rest $2$ digits can be anything among remaining $5$ digits (including the digit $5$) $\implies 5 \times 5 \times 4 = 80$. That's the total number of 3 digit even numbers formed by these digits. But we also counted $3$-digit numbers starting with $0$ or $8$, count the number of such integers and subtract.

Answer 2

Even Number will end with $\left \{ 0,2,4,6 \text{ or }8 \right \}$.

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Let us divide the possiblities of even number with number less than $700$ in $3$ parts.

• Ending with $0$
• Ending with $\left \{ 2,4 \text{ or }6 \right \}$.
• Ending with $8$.

• Ending with $0$

$3$ digit number possible$= \underbrace{4}_{\text{4 choices among 2,4,5,6}} \times \underbrace{4}_{\text{4 choices among 2,4,5,6,8}} \times 1$

• Ending with $\left \{ 2,4 \text{ or }6 \right \}= \underbrace{3}_{\text{3 choices among 2,4,5,6}} \times \underbrace{4}_{\text{4 choices among 0,2,4,5,6,8}} \times \underbrace{3}_{\text{3 choices among 2,4,6}}$

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• Ending with $8$ $= \underbrace{4}_{\text{4 choices among 2,4,5,6}} \times \underbrace{4}_{\text{4 choices among 0,2,4,5,6}} \times 1$

$=16+36+16=68$

Answer 3

The first digit can be either 2,4,5 or 6

The second digit can be any of the digits

The last digit can be any of the digits except 5

Case 1: When first digit is 5. $$Number\ of\ ways\ = 1*5*4=20$$ Case 2: When first digit is not 5 and second digit is 5. $$Number\ of\ ways\ = 3*1*4=12$$

Case 3: When neither first not second digit is 5. $$Number\ of\ ways\ = 3*4*3=36$$

Total number of ways= $36+12+20=68$