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

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.

Using the digits 1, 2, 3 and 4 without repeating them, we can form 24 three digit numbers. The probability of getting numbers with the digit 4 in hundreds place and the digit 2 in tens place is _______

1. 1/12 , 12.5%
2. 1/2, 50%
3.  \(1/3 , \ 33\dfrac{1}{3}\)% (thirty three one third)
4. 1/5 , 20%

Answer (Detailed Solution Below)

Option 1 : 1/12 , 12.5%

Free

MP Police Constable: Memory Based Paper: 8 Jan 2022 Shift 1

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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How many 2

How many two digit numbers can be formed from the digits 1,2 3 4 and 5 if repetitions are allowed?

How many two digit numbers can be generated using the digits 1,2 3.4 5 without repeating any digit?

How many numbers of 4 digits can be formed using the digits 1,2 3 4 5 such that repetition of digits are allowed?