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`        The total number of 5 digits number of different digits in which the digit in the middle is the largest`
3 years ago

User
6 Points
```							case 1 :- number does not contain 0.from 1, 2, 3, 4, 5, 6, 7, 8, 9choose 5 digits by $\binom{9}{5}$ ways.one number is largest which occupies middle place. arrange rest four numbers in 4! ways. thus$\binom{9}{5}$ X 4! =3024.case 2 :- contains 0choose other four numbers from 1,2,3,4,5,6,7,8,9 in $\binom{9}{4 }$ ways. largest occupies third position.no of ways numbers can be arranged = 3 X 3 X 2 X 1 =18ways = $\binom{9}{4 }$ X 18 = 2268therefore our answer is 3024 + 2268 = 5292
```
3 years ago
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• 101 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions