How many combinations can you make with numbers \[1,2,3\]?
Answer
564.9k+ views
Hint:
In the given question, we have been asked how many combinations of different numbers are possible with the three given digits. Now, we have to calculate this answer for two cases – when repetition is allowed and, when repetition is not allowed.
Complete step by step solution:
We have to calculate the number of combinations possible with the digits \[1,2,3\].
Case 1 – when repetition is allowed
Let the three crosses in \[ \otimes \otimes \otimes \] represent the spaces for the three digits.
Now, the first place has three options, the second place has three options and the third place also has three options.
So, the total number of ways is, \[N = 3 \times 3 \times 3 = 27\]
Hence, when repetition is allowed, the number of ways is \[27\].
Case 2 – when repetition is allowed
Let the three crosses in \[ \otimes \otimes \otimes \] represent the spaces for the three digits.
Now, the first place can be either of the three numbers, the second place can be either of the remaining two numbers and the third place has just the one left number.
So, the total number of ways is, \[N = 3 \times 2 \times 1 = 6\]
Hence, when repetition is allowed, the number of ways is \[6\].
Note:
So, for solving questions of such type, we first write what has been given to us. Then we write down what we have to find. Then we write the formula which connects the two things. When we are not given any information about the repetition of digits, we consider both of the cases and find their answers separately.
In the given question, we have been asked how many combinations of different numbers are possible with the three given digits. Now, we have to calculate this answer for two cases – when repetition is allowed and, when repetition is not allowed.
Complete step by step solution:
We have to calculate the number of combinations possible with the digits \[1,2,3\].
Case 1 – when repetition is allowed
Let the three crosses in \[ \otimes \otimes \otimes \] represent the spaces for the three digits.
Now, the first place has three options, the second place has three options and the third place also has three options.
So, the total number of ways is, \[N = 3 \times 3 \times 3 = 27\]
Hence, when repetition is allowed, the number of ways is \[27\].
Case 2 – when repetition is allowed
Let the three crosses in \[ \otimes \otimes \otimes \] represent the spaces for the three digits.
Now, the first place can be either of the three numbers, the second place can be either of the remaining two numbers and the third place has just the one left number.
So, the total number of ways is, \[N = 3 \times 2 \times 1 = 6\]
Hence, when repetition is allowed, the number of ways is \[6\].
Note:
So, for solving questions of such type, we first write what has been given to us. Then we write down what we have to find. Then we write the formula which connects the two things. When we are not given any information about the repetition of digits, we consider both of the cases and find their answers separately.
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