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.
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Solution
The correct option is D 6
Given digits are 3, 5 and 7.
Distinct numbers can be formed by shifting the digits as shown below.
| Number formed | Hundreds place | Tens place | Units place |
| 753 | 7 | 5 | 3 |
| 735 | 7 | 3 | 5 |
| 573 | 5 | 7 | 3 |
| 537 | 5 | 3 | 7 |
| 375 | 3 | 7 | 5 |
| 357 | 3 | 5 | 7 |
So, a total of 6 distinct numbers can be formed using these digits.