Solution:
Nội dung chính
- 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 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?
[i] When repetition of digits is allowed:
No. of ways of choosing first digit = 5
No. of ways of choosing second digit = 5
No. of ways of choosing third digit = 5
By fundamental counting principle,
Total possible number of ways = 5×5×5 = 125.
[ii] When repetition of digits is not allowed:
No. of ways of choosing first digit = 5
No. of ways of choosing second digit = 4
No. of ways of choosing to third digit = 3
By
fundamental counting principle,
Total possible number of ways = 5×4×3 = 60
NCERT Solutions Class 11 Maths Chapter 7 Exercise 7.1 Question 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?
Summary:
[i] Total number of 3 digit numbers formed when repetition is allowed = 125, [ii] Total number of 3 digit numbers formed when repetition is not allowed = 60
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 3
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. Let 3-digit number be XYZ.
How many 3
Solution : [i] When repetition of digits is allowed:
No. of ways of choosing firsy digits = 5
No. of ways of choosing second digit = 5
No. of ways of choosing third digit = 5
Therefore, total possible numbers `= 5 xx 5 xx 5 = 125`
[ii] When repetition of digits is not allowed:
No.
How many 3
I was able to get this question, by changing 2, 3, 4 and 5 to 1, 2, 3 and 4; then multiplying 4 by 3 by 2 to give 24 possibilities.
How many 3
How many 3-digit numbers can be formed from the digits 1, 2, 3, 4 and 5 assuming that repetition of the digits is not allowed. = m x n x p = 5 x 4 x 3 = 60. How many 3-digit numbers can be formed from the digits 1, 2, 3, 4 and 5 assuming that repetition of the digits is allowed. Number of places for the digits = 3.
How many 3
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 3
No. of ways of choosing firsy digits = 5
No. of ways of choosing second digit = 5
No. of ways of choosing third digit = 5
Therefore, total possible numbers `= 5 xx 5 xx 5 = 125`
[ii] When repetition of digits is not allowed:
No.