How many two digit numbers can you make using the digits 1, 2, 3 and 4 without repeating the digits

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100 Questions 100 Marks 120 Mins

Given:

1,2,3,4 total number

Concept used:

Formula for permutations and combinations

ncr , nPr

Calculation:                                                                                        

ncr = (4c3) × (3c2) ×(2c1)

⇒ 4 × 3 × 2 = 24

When we fix two number 4, 2,

Two number pair are possible,।

(421,423)

Probability = 2 ÷ 24

⇒ 1 ÷ 12

∴ the probability of getting numbers with the digit 4 in hundreds place and the digit 2 in tens place is 1 ÷ 12

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Índice

• The correct option is A25The 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 formed from the digits 1,2 3 4 and 5 if repetitions are allowed?
• How many numbers of 3 digits can be formed with the digits 1,2 3 4 and 5 without any repetition of digits?
• How many two digit numbers can we make using the same digit repeated?
• How many 3 numbers can be formed from the digits 1,2 3 4 and assuming that I repetition of the digits is allowed II repetition of the digits is not allowed?

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Solution

The correct option is A25The 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 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.

∴ The number of 2-digit numbers formed from the given set with repetition =5P2+5=20+5=25.

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

∴ 9 possible two-digit numbers can be formed.

So, required number of ways in which three digit numbers can be formed from the given digits is 5×4×3=60.