How many 3 digit numbers can be made from the digits 0 to 4 if repetition is allowed?

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

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

How many 3