# 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? [closed]

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.

## closed as off-topic by user296602, user99914, N. F. Taussig, Namaste, Rhys SteeleJul 4 '18 at 20:57

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• What have you done so far? Do you have an answer which you want to double check? Please add your working to the question always, since it helps people answer your question better. If you have no working to show, then try to form some three digit numbers from the set, and see if you get any pattern. – астон вілла олоф мэллбэрг Jul 4 '18 at 5:12
• The very first thing is to choose the first digit (the hundreds digit). Because it cannot be zero, the only options left are $2, 4, 5,6$. Surely, you can continue from here. – Matti P. Jul 4 '18 at 5:15

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

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

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

$=16+36+16=68$

• In your second case, you wrote $4$ choices among $2, 4, 5, 6$ but wrote $3$ for the number of choices. The number of choices for the hundreds digit depends on whether or not $8$ is the units digit. – N. F. Taussig Jul 4 '18 at 7:38
• @N.F.Taussig sir.Sorry it was a typo. – laura Jul 4 '18 at 7:46
• for the second case, i am first selecting units place by $4$ ways and then selecting the thousand's pace by $3$ ways$– laura Jul 4 '18 at 7:47 • If the units digit is$2$,$4$, or$6$, then$8$is a potential tens digit. The other two cases are correct. – N. F. Taussig Jul 4 '18 at 8:35 • You have a typographical error for the middle case. You meant to multiply by$4$choices for the tens digit. Otherwise, everything is correct. – N. F. Taussig Jul 4 '18 at 9:05 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=$36+12+20=68$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. • We want the number of$3$-digit even numbers less than$700\$ that can be formed from the given digits without repetition. Why are you adding the one-digit and two-digit numbers to the total? – N. F. Taussig Jul 4 '18 at 7:40
• @N.F.Taussig yes, there is no need for that – ab123 Jul 4 '18 at 10:56