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Total number of five digit numbers in which every succeeding digit is greater than the preceding digit is:
ans:9!/4!*5!
plz explain the steps that u would use!
Hi Basit,
Here basically we are forming a five digit number in ascending order.
So 0 cannot be there in any place (Conclusion 1).
So based on that conclusion, we have 9 digits {1,2,3,4,.....9} to pick from.
See from this we pick 5 numbers in 9C5 ways. Now say suppose you have picked a particular set of 5 numbers from the above 9 numbers..... after picking, those 5 numbers can be arranged in ascending order in only one way from the lowest to the greatest. (Say suppose you had picked 7,3,5,8,2..... they can be arranged in Asc Ord only like 2,3,5,7,8). Similarly for all the 9C5 sets you pick, can arrange in Asc Ord in only 1 way.
So we can do this in 9C5 ways. = the above answer.
Best Regards,
Ashwin (IIT Madras).
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