0 votes 0 votes How many 5 digit numbers can be formed if either (a) every successive digit exceeds its predecessor ,or (b) every successive digit is smaller than it’s predecessor Note:leading 0’s don’t count ,like 04521 is not a 5-digit number._ Prateek Raghuvanshi asked Jan 9, 2019 Prateek Raghuvanshi 485 views answer comment Share Follow See all 3 Comments See all 3 3 Comments reply arvin commented Jan 9, 2019 reply Follow Share a)126.. b)252.. 1 votes 1 votes shreyansh jain commented Jan 9, 2019 reply Follow Share $^9C_5+^{10}C_5$? 1 votes 1 votes gauravkc commented Jan 9, 2019 reply Follow Share a) Strictly Increasing. Choose any 5 out of 9 and sort them (as 0 cannot come. Even if we include 0 it'll be at first place which is not allowed) = $^9C_5$ = 126 b) Strictly decreasing. Choose any 5 out of 10 and sort. If 0 is selected it'll be at last place = $^{10}C_5$ = 252 2 votes 2 votes Please log in or register to add a comment.
