How many 3 digit even numbers can be formed using the digits 1 2 3 4 5 repetition allowed?

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1) How many 3-digit numbers can be formed by using $0,1,2,3,4,5$ ? Using basics it would be $ 5 \times 5 \times4 = 100$

2) How many 3-digit numbers can be formed by $8,1,2,3,4,5$ which are even? Again using basics we get $ 4 \times 5 \times 3 =60$

3) Now I want to ask how many 3 digit numbers can be formed which are even using $0,1,2,3,4,5$?

No repetition is allowed in all above cases. Here I am not getting how to use basics when we need to apply both conditions of case 1 and case 2 (i.e when we need to take care of both things zero at hundredth place and even number at unit place) simultaneously .

asked Feb 23, 2018 at 19:14

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Without repetition counting the number of 3digit numbers using digits from $\{0,1,2,3,4,5\}$:

Count the number of $3$-digit strings whose last digit is even.

• Pick the last digit: $3$ options
• Pick the first digit: $5$ options
• Pick the second digit: $4$ options

$3\times 5\times 4 = 60$ total

Remove from that the number of $3$-digit strings whose last digit is even and first digit is $0$

• First digit is zero: $1$ option
• Pick the last digit: $2$ options
• Pick the second digit: $4$ options

$1\times 2\times 4=8$ total that were "bad" and should not have been counted in the first if we wanted to count numbers instead of strings

This gives $60-8=52~~~$ $3$-digit even numbers using digits from $\{0,1,2,3,4,5\}$ without repetition.

answered Feb 23, 2018 at 19:53

JMoravitzJMoravitz

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How many even 3 digits numbers can be formed, with no repetition, using 0,1,2,3,4,5 ?

\begin{array}{cccccr} E & E & E & \to 2 \times 2 \times 1 &= & 4 \\ E & O & E & \to 2 \times 3 \times 2 &= & 12 \\ O & E & E & \to 3 \times 3 \times 2 &= & 18 \\ O & O & E & \to 3 \times 2 \times 3 &= & 18 \\ \end{array}

Total number of ways = $52$

How many even 3 digits numbers can be formed, with repetition, using 0,1,2,3,4,5 ?

\begin{array}{cccccr} E & E & E & \to 2 \times 3 \times 3 &= & 18 \\ E & O & E & \to 2 \times 3 \times 3 &= & 18 \\ O & E & E & \to 3 \times 3 \times 3 &= & 27 \\ O & O & E & \to 3 \times 3 \times 3 &= & 27 \\ \end{array}

Total number of ways = $90$

answered Feb 23, 2018 at 19:38

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We want 1, 2, 3, 4 or 5 in the hundreds digit - 5 choices, anything in the tens digit - 6 choices, and 0, 2 or 4 in the units digit - 3 choices. We have 5*6*3=90 numbers at the moment.

Now for the no repetition condition:

First we exclude odd digit-repeated odd digit-even digit numbers, 9 of these; and odd digit-even digit-repeated even digit numbers, another 9; also we exclude even digit-odd digit-repeated even digit numbers, 6 of these. For the all even digited numbers 2*2*1=4 out of the 2*3*3=18 do not have repetition so we exclude another 14.

We excluded 9+9+6+14=38 numbers. Hence our final solution is 90-38=52 numbers.

answered Feb 23, 2018 at 19:26

pilgrimpilgrim

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3

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