How many 3-digit numbers can be formed by using the digits 1 to 9 if no digit is repeated?
Solution and Explanation
3-digit numbers have to be formed using the digits 1 to 9.
Here, the order of the digits matters.
Therefore, there will be as many 3-digit numbers as there are permutations of 9 different digits taken 3 at a time.
Therefore, required number of 3-digit numbers=\(^9P_3\)
\(=\frac{9!}{(9-3)!}\)
\(=\frac{9!}{6!}\)
\(=\frac{9\times8\times7\times6}{6!}\)
\(=9\times8\times7=504\)
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Concepts Used:
Permutations and Combinations
Permutation:
Permutation is the method or the act of arranging members of a set into an order or a sequence.
- In the process of rearranging the numbers, subsets of sets are created to determine all possible arrangement sequences of a single data point.
- A permutation is used in many events of daily life. It is used for a list of data where the data order matters.
Combination:
Combination is the method of forming subsets by selecting data from a larger set in a way that the selection order does not matter.
- Combination refers to the combination of about n things taken k at a time without any repetition.
- The combination is used for a group of data where the order of data does not matter.