How many 3 digit even numbers can be formed using the digits 1 2 3 4 5 Repetition not allowed?

How many 3-digit numbers can be formed from the digits 1, 2, 3, 4 and 5 assuming that:1) Repetition is not allowed.2) Repetition is allowed.

Answer

Hint: Fundamental principle of counting: According to the fundamental principle of counting if a task can be done in “m” ways and another task can be done in “n” ways, then the number of ways in which both the tasks can be done in mn ways.

Complete step by step solution:
Case [1]: Repetition not allowed:
The number of ways in which the ones place can be filled = 5 ways.
The number of ways in which the tens place can be filled = 4 ways (because repetition is not allowed. So, the choice of one place cannot be used).
 The number of ways in which hundreds place can be filled = 3 ways.
Hence according to the fundamental principle of counting the number of ways in which the three places can be filled to form a three-digit number = $5\times 4\times 3=60$
Hence the number of 3-digit numbers formed using the digits 1, 2, 3, 4 and 5 = 60 i.e 5! ways
Case [2]: Repetition is allowed:
The number of ways in which the ones place can be filled = 5 ways.
The number of ways in which the tens place can be filled = 5 ways (because repetition is allowed. So, the choice of one's place can be used).
 The number of ways in which hundreds place can be filled = 5 ways.
Hence according to the fundamental principle of counting the number of ways in which the three places can be filled to form a three-digit number = $5\times 5\times 5=125$
Hence the number of 3-digit numbers formed using the digits 1, 2, 3, 4 and 5 = 125

Note: The number of 3-digit numbers formed using the digits 1, 2, 3, 4 and 5 when repetition is not allowed is equivalent to the number of 3 Letter permutations of 5 distinct letter = ${}^{5}{{P}_{3}}=\dfrac{5!}{\left( 5-3 \right)!}=\dfrac{5!}{2!}=\dfrac{120}{2}=60$ which is the same result as above.

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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

5532 silver badges8 bronze badges

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3

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How many even three digit num bers can be formed from the dig its 1/2 5/6 and 9 without re peating any of the digits?