How Many 9 Digit Numbers Of Different Digits Can Be Formed? - Vedantu

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AnswerQuestion Answers for Class 12Class 12 BiologyClass 12 ChemistryClass 12 EnglishClass 12 MathsClass 12 PhysicsClass 12 Social ScienceClass 12 Business StudiesClass 12 EconomicsQuestion Answers for Class 11Class 11 EconomicsClass 11 Computer ScienceClass 11 BiologyClass 11 ChemistryClass 11 EnglishClass 11 MathsClass 11 PhysicsClass 11 Social ScienceClass 11 AccountancyClass 11 Business StudiesQuestion Answers for Class 10Class 10 ScienceClass 10 EnglishClass 10 MathsClass 10 Social ScienceClass 10 General KnowledgeQuestion Answers for Class 9Class 9 General KnowledgeClass 9 ScienceClass 9 EnglishClass 9 MathsClass 9 Social ScienceQuestion Answers for Class 8Class 8 ScienceClass 8 EnglishClass 8 MathsClass 8 Social ScienceQuestion Answers for Class 7Class 7 ScienceClass 7 EnglishClass 7 MathsClass 7 Social ScienceQuestion Answers for Class 6Class 6 ScienceClass 6 EnglishClass 6 MathsClass 6 Social ScienceQuestion Answers for Class 5Class 5 ScienceClass 5 EnglishClass 5 MathsClass 5 Social ScienceQuestion Answers for Class 4Class 4 ScienceClass 4 EnglishClass 4 MathsHow many 9 digit numbers of different digits can be formed?AnswerVerified591.3k+ viewsHint: We will use the concept of the relation between permutation and combination. This is given by the formula $P_{r}^{n}=r!C_{r}^{n}$ where $C_{r}^{n}=\dfrac{n!}{r!\left( n-r \right)!}$. Here n is representing the total number of objects and r is denoting selected items. We will use this relation in order to solve the question.Complete step-by-step answer:According to the question we have a total number of digits as 10. This is because we can form any number by using the digits 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9 which are 10 in total. We are going to find 9 digit numbers with different digits. We will fill up the first place by all the 9 digits and not 10. This is because if zero comes in any 9 digit number and zero is included in it then it will become an 8 digit number instead of 9. Therefore, the first place is filled by 9 digits only or we can say that it will have 9 choices only. Now the rest of the places can be filled by using the relation of permutation and combination here. We will use the formula $P_{r}^{n}=r!C_{r}^{n}$which results into $\begin{align} & P_{r}^{n}=r!C_{r}^{n} \\ & \Rightarrow P_{r}^{n}=r!\dfrac{n!}{r!\left( n-r \right)!} \\ & \Rightarrow P_{r}^{n}=\dfrac{n!}{\left( n-r \right)!} \\ \end{align}$Here n is representing the total number of objects and r is denoting selected items. As here we have n = 9 and r = 8 therefore, we have$\begin{align} & P_{8}^{9}=\dfrac{9!}{\left( 9-8 \right)!} \\ & \Rightarrow P_{8}^{9}=\dfrac{9!}{1!} \\ & \Rightarrow P_{8}^{9}=9! \\ \end{align}$ Therefore, the total number of possibilities is given by $9\times 9!=3265920$.Hence, 9 digit numbers of different digits can be formed in 3265920 ways.Note: Alternatively we can use different methods in which we will use the concept of finding out the choices for each place. This is done below. We can form any number by 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9 digits. In order to form a 9 digit number the first digit should be free from 0. Therefore, for the first place we have 9 choices only. So we now have $9,-,-,-,-,-,-,-,-$. We can consider the other places that can be filled by any number also, we can consider 0 for them. But in this question the limitation given to us is that the repetition is not allowed. Therefore, as we have a total 10 digits given by 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9 and out of these one is already used in the first place. Therefore, we are left with now 9 choices. That is $9,9,-,-,-,-,-,-,-$. And the third place has 8 choices as there is no repetition. That is $9,9,8,-,-,-,-,-,-$. Continuing in this manner we will get $9,9,8,7,6,5,4,3,2$. Therefore, by multiplying the choices we have a total number of possibilities given by $9\times 9\times 8\times 7\times 6\times 5\times 4\times 3\times 2=3265920$. Hence, 9 digit numbers of different digits can be formed in 3265920 ways.Recently Updated PagesMaster Class 12 Business Studies: Engaging Questions & Answers for SuccessMaster Class 12 Economics: Engaging Questions & Answers for SuccessMaster Class 12 English: Engaging Questions & Answers for SuccessMaster Class 12 Maths: Engaging Questions & Answers for SuccessMaster Class 12 Social Science: Engaging Questions & Answers for SuccessMaster Class 12 Chemistry: Engaging Questions & Answers for SuccessMaster Class 12 Business Studies: Engaging Questions & Answers for SuccessMaster Class 12 Economics: Engaging Questions & Answers for SuccessMaster Class 12 English: Engaging Questions & Answers for SuccessMaster Class 12 Maths: Engaging Questions & Answers for SuccessMaster Class 12 Social Science: Engaging Questions & Answers for SuccessMaster Class 12 Chemistry: Engaging Questions & Answers for Success

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