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The given digits are 1, 2, 3, 4, 5

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    The correct option is A25 The given number of digits is 6. The tenths place of the two digit number can be filled by 5 digits (0 cannot be used to fill this place). The units place of the two digit number can be filled by the other four digits and zero also.So the number of 2 digit numbers that can be formed using 0, 1, 2,3,4,5 is 5x5 is 25.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 two digit numbers can be generated using the digits 1.2 3.4 5 without repeating any digit?How many 2 digit numbers can be formed by using the digits 3 5 6 without repeating any digit?How many numbers can be formed from 12345 without repetition?How many possible two digit numbers can be formed by using the digits 2 5 and 9 repetition of digits is allowed )?

A two digit number has unit place and 10’s place.

We are given 5 digits (1, 2, 3, 4, 5).

The unit place can be filled (using the 5 digits) in 5 ways.

After filling the unit place since repetition is not allowed one number (filled in the unit place) should be excluded.

So the 10’s place can be filled (using the remaining 4 digits) in 4 ways

∴ Unit place and 10’s place together can be filled in 5 × 4 = 20 ways.

So the number of two-digit numbers = 20

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Solution

The correct option is A25 The given number of digits is 6. The tenths place of the two digit number can be filled by 5 digits (0 cannot be used to fill this place). The units place of the two digit number can be filled by the other four digits and zero also.So the number of 2 digit numbers that can be formed using 0, 1, 2,3,4,5 is 5x5 is 25.

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 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 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 2nd place = 9 (As one letter has already been chosen 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 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 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 two digit numbers can be generated using the digits 1.2 3.4 5 without repeating any digit?

Total Number of Numbers which can be formed by numbers 1,2,3,4,5 (without repeating digitsi) = 5*4*3*2*! = 5! = 120.

How many 2 digit numbers can be formed by using the digits 3 5 6 without repeating any digit?

Detailed Solution Hence 12, 2-digit & 3-digit numbers can be formed by using the digits 3, 5, 6 without repeating any digit.

How many numbers can be formed from 12345 without repetition?

Total = 60. (i) the ten's place (ii) the unit's place?

How many possible two digit numbers can be formed by using the digits 2 5 and 9 repetition of digits is allowed )?

Detailed Solution ∴ 9 possible two-digit numbers can be formed. Tải thêm tài liệu liên quan đến nội dung bài viết How many two digit numbers can be generated using the digits 1,2 3.4 5 without repeating any digit?

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