How many numbers can be formed with odd digits 1, 3, 5, 7, 9 without repetition

Using the digits 1, 2, 3, 5, and 6, without repetition, how many five-digit even numbers can be formed?

1. 51
2. 24
3. 48
4. 96
5. 64

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• Using the digits 1, 2, 3, 5, and 6, without repetition, how many five-digit even numbers can be formed?
• 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 numbers can be formed using the digits 1 3 4 5 6 8 and 9 if I no repetitions are allowed II repetitions are allowed?
• How many 3 digits number can be formed using the digits 1 3 5 7 9 where we are allowed to repeat the digits?

Concept:

Basic Principle of Counting:

If there are m ways for happening of an event A, and corresponding to each possibility there are n ways for happening of event B, then the total number of different possibilities for happening of events A and B are:

• Either event A OR event B alone = m + n.
• Both event A AND event B together = m × n.
• An even number's units digit is either 0, 2, 4, 6 or 8.

Calculation:

For an even number, the units place must be 2 or 6, so it can be filled in 2 ways.

The remaining four places can now be filled in 4, 3, 2 and 1 ways respectively.

The total number of ways in which the number can be written = 4 × 3 × 2 × 1 × 2 = 48.

a) repetitions are allowed; b) repetitions not allowed

My solution: a) There are 3 cases:

1. $3$-digit numbers;
2. $2$-digit numbers;
3. one digit numbers.

Case 1: $3\times 5\times5=75$

Case 2: $5\times5=25$

Case 3: $5 \times$ Sum of three cases $= 105$. (ans. key says $2253$?)

b) no repetitions:

Case 1: $3\times 4\times 3=36$

Case 2: $5\times 4=20$

Case 3: $5 \times$ Sum of cases $= 61$ (ans. key says $195$)

My answers are way different. Not sure what I'm missing. Help would be appreciated.

Answer : 64

Solution : One-digit numbers:
Clearly, there are four 1 -digit numbers.
Two-digit numbers:
We may fill the unit's place by any of the four given digits.
Thus, there are 4 ways to fill the unit's place.
The ten's place may now be filled by any of the remaining three digits. So, there are 3 ways to fill the ten's place.
Number of 2-digit numbers `=(4xx3)=12.`
Three-digit numbers:
Number of ways to fill the unit's, ten's and hundred's places are 4, 3 and 2 respectively.
Number of 3-digit numbers `=(4xx3xx2)=24.`
Four-digit numbers:
Number of ways to fill the unit's ten's, hundred's and thousand's places are 4, 3, 2 and 1 respectively.
Number of 4-digit numbers `=(4xx3xx2xx1)=24.`
Hence, the number of required numbers `=(4+12+24+24)=64.`

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 numbers can be formed using the digits 1 3 4 5 6 8 and 9 if I no repetitions are allowed II repetitions are allowed?

Hence, 64 numbers can be formed.

How many 3 digits number can be formed using the digits 1 3 5 7 9 where we are allowed to repeat the digits?

Therefore, a total of 100 3 digit numbers can be formed using the digits 0, 1, 3, 5, 7 when repetition is allowed.

How many 4

Expert-verified answer Thus 24 numbers can be formed. As all digits are unique answer is 24 4-digit no. Can be formed.

How many 4

nAnswer: 840" Was this answer helpful?

How many 3

How many 3

How many 3

How many ways the three digits 1/2 and 3 can be arranged if repetition is allowed?