SATHEE: Permutations And Combinations Question 250

Permutations And Combinations Question 250

Question: Find the total number of 9 digit numbers which have all the digits different [IIT 1982]

Options:

A) $ 9\times 9\ ! $

B) $ 9\ ! $

C) 10!

D) None of these

Show Answer

Answer:

Correct Answer: A

Solution:

There are 10 digits in all viz. 0, 1, 2, 3, 4, 5, 6, 7, 8, 9. The required 9 digit numbers = (Total number of 9 digit numbers including those numbers which have 0 at the first place) - (Total number of those 9 digit numbers which have 0 at the first place) $ ={{}^{10}}P_9-{{}^{9}}P_8=\frac{10!}{1!}-\frac{9!}{1!}=10!-9! $ = $ (10-1)9!=9.9! $ .

BETA

sathee@iitk.ac.in

Media coverage of SATHEE

Play Store

App Store

The Ministry of Education has launched the SATHEE initiative in association with IIT Kanpur to provide free guidance for competitive exams. SATHEE offers a range of resources, including reference video lectures, mock tests, and other resources to support your preparation. Please note that participation in the SATHEE program does not guarantee clearing any exam or admission to any institute.

Ask SATHEE

Welcome to SATHEE !
Select from 'Menu' to explore our services, or ask SATHEE to get started. Let's embark on this journey of growth together! 🌐📚🚀🎓

I'm relatively new and can sometimes make mistakes.
If you notice any error, such as an incorrect solution, please use the thumbs down icon to aid my learning.
To begin your journey now, click on I understand

Please select your preferred language
कृपया अपनी पसंदीदा भाषा चुनें

Permutations And Combinations Question 250

Question: Find the total number of 9 digit numbers which have all the digits different [IIT 1982]

Options:

Answer:

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Mentorship

NCERT Video Solution

Menu