How many 3 digits number can be formed from the digits 2 3 4 5 6 Assuming that a repetition of the digits is allowed B repetition is not allowed?

(a) How many three-digit numbers can be formed from the digits 0, 1, 2, 3, 4, 5, and 6 if each digit can be used only once? (b) How many of these are odd numbers? (c) How many are greater than 330?

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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 numbers can be formed using the digits 2 3 4 5 6 if digits can be repeated?
• How many 3 digits numbers can be formed the digits 1 2 3 4 and 5 assuming I repetition of digits allowed II repetition of digits not allowed?
• How many 2 digit & 3

This is what I have done:

a) We can choose from 6 numbers for the first digit ( we exclude 0), 6 digits for the second (we exclude the first but include 0) and finally 5 digits for the third (we exclude the first and second). So total number of$$ \text{possibilities} = 6 \cdot 6 \cdot 5 = 180 \text{ways}$$

b) I have no idea how to approach this. How can we do this?

c) I considered the case when the first digit is 3 , then for the second digit we have the possibilities of {4,5,6} and the last digit {0,1,3,4,5,6}. However we exclude 3, and one more number that has been chosen as the 2nd digit for our last number. So the number of $$\text{possibilities} = 1\cdot 3 \cdot (7-2) = 15$$

Now I considered when the first digit is greater than 3, {4,5,6} then for the second digit we can use {0,1,2,3,4,5,6} (but we exclude the number that has been used as the first digit). Finally for the third {0,1,2,3,4,5,6} and we exclude 2 numbers than have been used. So the number of $$ \text{possibilities} = 3 \cdot (7-1) \cdot (7-2) =90$$

In conclusion we have: $$90 + 15 = 105$$ total possibilities greater than 330.

Thank you for your time!

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 numbers can be formed using the digits 2 3 4 5 6 if digits can be repeated?

How many 3 digits numbers can be formed the digits 1 2 3 4 and 5 assuming I repetition of digits allowed II repetition of digits not allowed?

How many 3

How many 2 digit & 3