How many 3 digit numbers can be formed using 1, 2, 3, 4, 5 without repetition

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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.

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.

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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`.

Solution

(i) The unit place can be filled by any one of the digits 1,2,3,4 and 5. So the unit place can be filled in 5 ways. Similarly the tens place and hundreds place can be filled in 5 ways each because repetition of digits is allowed.∴ Total number of 3-digit numbers = 5×5×5=125(ii) The unit place can be filled by any one of the digits 1,2,3,4 and 5. So the unit place can be filled in 5 ways. Now four digits are left. So the tens place can be filled in 4 ways. the hundreds place can be filled in 3 ways by the remaining 3 digits because repetition of digits is not allowed.∴ Total number of 3-digit numbers = 3×4×5=60.

Problem 1: How many 3-digit numbers can be formed from the digits 1, 2, 3, 4 and 5 assuming that – 

(i) repetition of the digits is allowed?

Solution: 

Answer: 125.

Method:

Here, Total number of digits = 5

Let 3-digit number be XYZ.

Now the number of digits available for X = 5,

As repetition is allowed,

So the number of digits available for Y and Z will also be 5 (each).

Thus, The total number of 3-digit numbers that can be formed = 5×5×5 = 125.

(ii) repetition of the digits is not allowed? 

Solution:

Answer: 60.

Method:

Here, Total number of digits = 5

Let 3-digit number be XYZ.

Now the number of digits available for X = 5,

As repetition is not allowed,

So the number of digits available for Y = 4 (As one digit has already been chosen at X),

Similarly, the number of digits available for Z = 3.

Thus, The total number of 3-digit numbers that can be formed = 5×4×3 = 60.

Problem 2: How many 3-digit even numbers can be formed from the digits 1, 2, 3, 4, 5, 6 if the digits can be repeated?

Solution:

Answer: 108.

Method:

Here, Total number of digits = 6

Let 3-digit number be XYZ.

Now, as the number should be even so the digits at unit place must be even, so number of digits available for Z = 3 (As 2,4,6 are even digits here),

As the repetition is allowed,

So the number of digits available for X = 6,

Similarly, the number of digits available for Y = 6.

Thus, The total number of 3-digit even numbers that can be formed = 6×6×3 = 108.

Problem 3: How many 4-letter code can be formed using the first 10 letters of the English alphabet if no letter can be repeated? 

Solution:

Answer: 5040

Method:

Here, Total number of letters = 10

Let the 4-letter code be 1234.

Now, the number of letters available for 1st place = 10,

As repetition is not allowed,

So the number of letters possible at 2nd place = 9 (As one letter has already been chosen at 1st place),

Similarly, the number of letters available for 3rd place = 8,

and the number of letters available for 4th place = 7.

Thus, The total number of 4-letter code that can be formed = 10×9×8×7 = 5040.

Problem 4: How many 5-digit telephone numbers can be constructed using the digits 0 to 9 if each number starts with 67 and no digit appears more than once?

Solution:

Answer: 336

Method:

Here, Total number of digits = 10 (from 0 to 9)

Let 5-digit number be ABCDE.

Now, As the number should start from  67 so the number of possible digits at A and B = 1 (each),

As repetition is not allowed,

So the number of digits available for C = 8 ( As 2 digits have already been chosen at A and B),

Similarly, the number of digits available for D = 7,

and the number of digits available for E = 6.

Thus, The total number of 5-digit telephone numbers that can be formed = 1×1×8×7×6 = 336.

Problem 5: A coin is tossed 3 times and the outcomes are recorded. How many possible outcomes are there?

Solution:

Answer: 8

Method:

We know that, the possible outcome after tossing a coin is either head or tail (2 outcomes),

Here, a coin is tossed 3 times and outcomes are recorded after each toss,

Thus, the total number of outcomes = 2×2×2 = 8.

Problem 6: Given 5 flags of different colours, how many different signals can be generated if each signal requires the use of 2 flags, one below the other?

Solution:

Answer: 20.

Method:

Here, Total number of flags = 5

As each signal requires 2 flag and signals should be different so repetition will not be allowed,

So, the number of flags possible for the upper place = 5,

and the number of flags possible for the lower place = 4.

Thus, the total number of different signals that can be generated = 5×4 = 20.

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