How many three digit numbers can be made from the digits 0 to 4 if repetition is allowed

centripetal question given that how many three digit numbers can be formed by using the digits 0 1 2 and 3 where the repetition is allowed ok repetition of digits are allowed and we need to find out how many 3 digit number can be formed using the digits 0 1 2 3 4 digit is given to us is the digits which are given to us are 0123 that means there are four digit given to us there are in total 4 digit given to us right and three digit number there are three places that support this is the first place this is the second place and this is the third place that made this is the ones place this is the ones place this is the tens place tens place and this is the hundred place if this is the hundred place rights to these are the three digits of a three digit number the festival will be start from the ones place

start from the ones place once place can accommodate how many digits to this one place ke comedy at either Zero OR one or two or three right from all the three all the four digit we can accommodate all the four digits that is 01234 the ways in which the ones place can be filled is for drive that can accommodate 4 digit similarly in the tens place is Kurnool all the four digits that is 0 1 2 and 3 all the four digits of the ways in which the tens place will be filled will also be equal to 4 but in the hundreds place this can only accommodate the digits 1 2 3 because if 0 will be there then this will not become a three digit number then this will become a two digit number because it will be over here then this place will not be counted then it will become only two digit number so in this particular place only 12 13 can be filled their ways can be this will be three three ways can be there to fill up the haunted place

so total numbers of 3 digit number will be equal to the total number of three digit number the total number of of 3 digit number 3 digit number will be equal to will be equal to 3 X the tens place that is 4 x the ones place that is again for that this will be equal to 3 into 16 that is 4 into 4 8 16 32 16 will be equal to 48 48 search 3 digit number can be formed in which the digits will be either 0123 the answer of the question that 48 Sach number can be found

The four given digits are 0, 1, 2 and 3.

Nội dung chính Show

• 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?
• (ii) repetition of the digits is not allowed? 
• 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?
• 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? 
• 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?
• Problem 5: A coin is tossed 3 times and the outcomes are recorded. How many possible outcomes are there?
• 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?
• How many three digit number can be formed using the digits 1,2 3 and 4 if repetitions are allowed?

Three-digit numbers are of the form

We need to find the number of 3-digit numbers that can be formed using the given digits without repetition.

0 cannot be at the hundreds place. Thus, the hundreds place can be filled with 3 digits (1, 2 or 3).

Then, the tens place can be filled with the remaining 3 digits and the ones place can be filled with the remaining 2 digits.

Thus, there are

Complete step-by-step solution:
In this question, we are asked how many 3 digit numbers can be formed using the digits 0, 1, 3, 5, 7 where repetition is allowed.
Given digits: 0, 1, 3, 5, 7
Therefore, there are a total of 5 digits.
Now, as we have to form 3 digit numbers, let us draw three boxes.

Therefore, there are 4 ways to fill the first box.
Now, for the second box, any out of the 5 given digits can be used to fill the box as repetition is allowed.
Therefore,

And for the third box too, we can fill it using any of the 5 given digits. Therefore, we get

So, therefore the total ways for forming 3 digit number will be
$ \Rightarrow $Total ways of arrangement $ = 4 \times 5 \times 5 = 100$
Therefore, a total of $100$ 3 digit numbers can be formed using the digits 0, 1, 3, 5, 7 when repetition is allowed.

Note: Here, note that if repetition were not allowed, then we would have got different answers.
For first digit:
As we have to form a 3 digit number, we cannot use 0 as the first digit. So,

For second digit:
Now, suppose we used 1 as the first digit, so now it cannot be used again for the second and third digit. So, the second digit can be 0, 3, 5 or 7. Therefore,

For third digit:
Suppose we used 3 as the second digit, so now 3 cannot be used for the third digit. So, the third digit can be 0, 5 or 7. Therefore,

$ \Rightarrow $Total numbers formed when repetition is not allowed$ = 4 \times 4 \times 3 = 48$

How many three-digit numbers can be formed from the digits 0, 1, 3, 5, 6 if repetitions of digits are not allowed?

Three-digit number is to be formed from the digits 0, 1, 3, 5, 6
When repetition of digits is not allowed:
100’s place digit should be a non zero number.
Hence, it can be anyone from digits 1, 3, 5, 6
∴ 100’s place digit can be selected in 4 ways.
0 can appear in 10’s and the unit’s place and digits can’t be repeated.
∴ 10’s place digit can be selected in 4 ways and the unit’s place digit can be selected in 3 ways
∴ By using the fundamental principle of multiplication, the total number of three-digit numbers = 4 × 4 × 3 = 48

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.

How many three digit number can be formed using the digits 1,2 3 and 4 if repetitions are allowed?

So the number of 3-number combinations are- (1,2,3),(1,2,4),(1,3,4),(2,3,4). Each can be arranged in 6 ways, so we get 24 ways totally.

How many 3

∴ Total number of 3-digit numbers = 3×4×5=60.

How many 3

(i) repetition of the digits is allowed? Solution: Answer: 125.

How many 3

ANSWER: 120 three-digit numbers can be formed WITHOUT REPETITION OF DIGITS.

How many 3

How many 3

How many 3

How many even 4 digit numbers are there if repetition is not allowed?