How many 3 digit numbers can you make using the digits 1/2 and 3 with repetitions?

Answer

Verified

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.

Solution : (i) When repetition of digits is allowed:
No. of ways of choosing firsy digits = 5
No. of ways of choosing second digit = 5
No. of ways of choosing third digit = 5
Therefore, total possible numbers `= 5 xx 5 xx 5 = 125`
(ii) When repetition of digits is not allowed:
No. of ways of choosing first digit = 5
No. of ways of choosing second digit = 4
No. of ways of choosing thrid digit = 3
Total possible numbers `= 5 xx 4 xx 3 = 60`.

Answer

Nội dung chính Show

  • How many 3 digit numbers can be formed using the numbers 6, 1, 2, 3 without repetition ?
  • Answer (Detailed Solution Below)
  • How many 3 digit numbers be formed using the digits 1/2 and 3 without repetition are divisible by 6?
  • How many combinations of 3 numbers can you have without repetition?
  • How many numbers are formed using the digits 1,2 3 without repetition?
  • How many 3 digit even numbers can be made by using the digits 1,2 3 4 6 if repetition of digits is not allowed?

Verified

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.

How many 3 digit numbers can be formed using the numbers 6, 1, 2, 3 without repetition ?

  1. 36
  2. 24
  3. 18
  4. 12

Answer (Detailed Solution Below)

Option 2 : 24

Free

Punjab Police SI 2016 Official Paper

100 Questions 100 Marks 90 Mins

Given:

Number are formed by using 6, 1, 2, 3

Calculation:

Number of ways to fill the first position = 4

Number of ways to fill second position = 3

Number of ways to fill third position = 2

Total number of ways = 4 × 3 × 2

⇒ Total number of ways = 24

∴ The total number of 3 digit number is 24.


Additional Information

If the repetition of number was allowed then we can take each digit in 4 different ways and so, the answer will be 4 × 4 × 4 = 64

Latest Punjab Police SI Updates

Last updated on Oct 12, 2022

The tentative answer key for the Punjab Police SI Exam conducted on 16th October 2022 has been released. Candidates can raise their objections till 22nd October 2022. The Punjab Police has released the Punjab Police SI Admit Card for the exam that will be conducted on 16th October 2022. The exam will have 2 papers. Paper 1 will be of 2 hours conducted from 9 am to 11 am and paper II of 2 hours conducted from 3 pm to 5 pm. Both exams will be 400 marks each. With the exam, a candidate also needs to pass the physical standard test to get selected for the Punjab Police SI post. For the Punjab police sub-inspector post, a total of 560 vacancies were released. The pay scale for a sub-inspector position is between Rs. 10300 and Rs. 35400.

Made by

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.

Solution : (i) When repetition of digits is allowed:
No. of ways of choosing firsy digits = 5
No. of ways of choosing second digit = 5
No. of ways of choosing third digit = 5
Therefore, total possible numbers `= 5 xx 5 xx 5 = 125`
(ii) When repetition of digits is not allowed:
No. of ways of choosing first digit = 5
No. of ways of choosing second digit = 4
No. of ways of choosing thrid digit = 3
Total possible numbers `= 5 xx 4 xx 3 = 60`.

Uh-Oh! That’s all you get for now.

We would love to personalise your learning journey. Sign Up to explore more.

Sign Up or Login

Skip for now

Uh-Oh! That’s all you get for now.

We would love to personalise your learning journey. Sign Up to explore more.

Sign Up or Login

Skip for now

Home

How many 3 digit numbers can be formed using the digit 0,1,2,3 without repetitions?

Question

How many 3 digit numbers be formed using the digits 1/2 and 3 without repetition are divisible by 6?

Hence, a total of 2(132 and 312) three-digit numbers can be formed using 1, 2, 3 without any repetition and which are also divisible by 6.

How many combinations of 3 numbers can you have without repetition?

There are 504 different 3-digit numbers which can be formed from numbers 1, 2, 3, 4, 5, 6, 7, 8, 9 if no repetition is allowed. Note: We can also use the multiplication principle to answer this question.

How many numbers are formed using the digits 1,2 3 without repetition?

Originally Answered: how many 3 digit numbers can be formed using 1,2 and 3 without repeating a digit ? Thus there are 6 possibilities.

How many 3 digit even numbers can be made by using the digits 1,2 3 4 6 if repetition of digits is not allowed?

Thus, by multiplication principle, the required number of 3-digit numbers is 3×20=60.

How many 3

Answer: 24 As repetition is not allowed, So the number of digits available for B = 3 (As one digit has already been chosen at A), Similarly, the number of digits available for C = 2.

How many combinations can you make with the numbers 1,2 3 with repetition?

That is a total of 7 combinations.

How many 3

So I take some particular numbers, like 1,2,3 and say that, well, 1 can go in 3 places, 2 in 2 places and 3 in 1 place, so by multiplication principle, there are 6 ways of forming a 3-digit number with 1,2,3. But there are 4 different numbers. So the number of 3-number combinations are- (1,2,3),(1,2,4),(1,3,4),(2,3,4).

How many 3

Hence, a total of 2(132 and 312) three-digit numbers can be formed using 1, 2, 3 without any repetition and which are also divisible by 6.