1 votes 1 votes Number of 5 digit number having there digits in non decreasing order (from left to right) constructed by using the digits belonging to the set {1, 2, 3, 4, 5, 6, 7, 8, 9} ? Combinatory discrete-mathematics combinatory + – Shubhanshu asked Apr 28, 2017 • retagged Jun 27, 2017 by Arjun Shubhanshu 740 views answer comment Share Follow See all 2 Comments See all 2 2 Comments reply Nirmal Gaur commented Apr 28, 2017 reply Follow Share it is a finite set, repetition is not possible 0 votes 0 votes Shubhanshu commented Apr 28, 2017 reply Follow Share Repetition is possible because it is not said that its digits are in increasing order. here it is non-decreasing order hence repetition is allowed. 0 votes 0 votes Please log in or register to add a comment.
2 votes 2 votes 5 digit no. with non decreasing order is same as selecting five digit and after selection we can arrange that 5 digit only in one way. if repetition is not allowed we can select in 9c5. and if repetition is allowed then we can select in 13c5. ANKUSH KUMAR answered Apr 28, 2017 ANKUSH KUMAR comment Share Follow See all 7 Comments See all 7 7 Comments reply Shubhanshu commented Apr 29, 2017 reply Follow Share If repetition is not allowed we can select in 9C5 ways that is perfect when we given that digits are in strictly in increasing order. But for repetition is allowed 13c5 how ??? 0 votes 0 votes ANKUSH KUMAR commented Apr 29, 2017 reply Follow Share if repetition is allowed then it is same as x1 + x2 + x3 + x4 + x5 + x6 + x7 + x8 + x9 = 5 where 0 <=xi <=5 . It is the case of 5 indistinguishable objects into 9 distinguishable boxes. $\binom{n+r-1}{r} = \binom{9+5-1}{5}$ 0 votes 0 votes Shubhanshu commented Apr 29, 2017 reply Follow Share why it is not the case where we have 5 boxes and 9 objects. since in the question also we have 5 places to fill with these 9 objects so the ways can be calculated as (9+5-1)C(5-1) = (9+5-1)C(9) as according to combinatorics (boxes + object - 1) C (boxes -1) = (boxes + object -1) C (objects) Correct me if I am wrong?? 0 votes 0 votes ANKUSH KUMAR commented Apr 29, 2017 reply Follow Share here we have x1 + x2 + x3 + x4 + x5 + x6 + x7 + x8 + x9 = 5 . for all xi ( box) we can leave the box vaccant ( xi = 0) and we can also take multiple value of every xi ( xi=1,2,3...). But the condition is total no of digit is 5. But in your case x1+x2+x3+x4+x5=9 does not make sense. you take places as xi which can not be vaccant and also we can not take multiple value for xi (box) . 1 votes 1 votes Shubhanshu commented Apr 29, 2017 reply Follow Share ok, I got it since in digit places we can place only one digit at a time. but in box and ball example we can put multiple balls at a time in a single box that's why it is calling as repetition and hence the number of boxes we have is 5 and objects as 9. but in our example, we have 9 boxes(digits from the set) and 5 balls(places in five-digit number). Through this, we can take the same digit for multiple places. ryt??? 0 votes 0 votes ANKUSH KUMAR commented Apr 29, 2017 reply Follow Share yes..u r ryt.. 0 votes 0 votes Shubhanshu commented Apr 29, 2017 reply Follow Share thanks .... 0 votes 0 votes Please log in or register to add a comment.