There are $N$ numbers of 11 digit positive integers such that the digits from left to right are non-decreasing, for example $12345678999,55555555555,23345557889$.
Find the sum of all digits of $N$.
Please help me out on this one.
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Kushagra Agarwal
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It seems to me you are saying that any 11 digit integer with no 0 is either decreasing or increasing. If yes, can you elaborate on why? – Calvin Lin Sep 08 '21 at 15:52
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Oops! That was a lame mistake. – Kushagra Agarwal Sep 08 '21 at 15:55
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Now that you've spotted the original mistake, can you try this again and add what you've tried (in a bit). If you're truly stuck, some of the related questions off the the side have relevant answers/ideas. – Calvin Lin Sep 08 '21 at 15:57
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I suggest starting with a smaller length than $11$. It should be easy for you to handle small lengths, and perhaps a pattern will emerge. – lulu Sep 08 '21 at 16:06
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does this answers your question? https://math.stackexchange.com/questions/1446797/finding-the-number-of-non-decreasing-sequences?rq=1 – atzlt Sep 08 '21 at 16:11
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Shouldn't this just be about finding the sum of all possible numbers , with repetition , because any selection can be adjusted accordingly . – Sukhoi234 Sep 08 '21 at 16:35
2 Answers
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HINT
In these types of problems, we should find how many number there are when the first digits is $1$ or $2$ etc. and find how many number there are when the second digits is $1$ or $2$, etc. You will find them up to $11$ digits.
Second problem is to find the number of possible arrangements. In this case, bijection helps us. The all thing we need to do is to select $11$ digits among $1,2,3,..,8,9$ when the repetition is allowed, the arrangement of number will be handled automatically. You can do it by the formula of combination with repetition.
By the way, notice that we do not take $0$ in our set, because if we take it, it will be always the first digit and it will not be an $11$-digit number.
For example, when the first digit is $1$, there are $10+9-1 \choose 10$ possible arrangements .
After that, find the summation of all digits , $10^{10} \times 18 \choose 10$ will give you the case where the first digits is $1$.
I think you can take it from here.
amWhy
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Not a Salmon Fish
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Can you please tell me how to find it for the first two digits so that I can get an idea about how to do for the rest? – Kushagra Agarwal Sep 09 '21 at 10:48
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1@KushagraAgarwal ,The main obstacle of this question is to find how many non-decreasing $11$ digits number there are , so we will follow the following procedure such that, lets assume that first digit is $1$ , then there are $10$ digits to arrange them in the order of non-decreasing such as $\color{blue}{1}1123456666$ or $\color{blue}{1}2334567889$ , right ? I say that if we select $10$ numbers in the set of ${1,2,..,8,9}$ , we can construct our $10$ digits non -decreasing number. SEE NEXT COMMENT.. – Not a Salmon Fish Sep 09 '21 at 11:57
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1@KushagraAgarwal To do it , we are selecting these $10$ number with repetiton ,i.e, we can select $8$ ten times , or we can select $5$ three times and $6$ two times and $9$ five times. We can make this selection by using combination with repetiton. Look at : https://sites.math.northwestern.edu/~mlerma/courses/cs310-05s/notes/dm-gcomb#:~:text=Any%20selection%20of%20r%20objects,at%20a%20time%20with%20repetition.&text=Two%20combinations%20with%20repetition%20are,times%2C%20regardless%20of%20their%20order. SEE NEXT COMMENT... – Not a Salmon Fish Sep 09 '21 at 11:58
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1@KushagraAgarwal For example , assume that we are trying to find how many number $abc$ are there such that $a \leq b \leq c$ , the answer is $C(9+3-1,3)$ ,realize that it is three digits number with non-decreasing order .You will find each digits in this way. However , realize that, for example, when the first digit is $3$ , you cannot use $1,2$ for the rest . – Not a Salmon Fish Sep 09 '21 at 11:58
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1@KushagraAgarwal I will put here very beneficial link for you : https://math.stackexchange.com/questions/3879521/find-the-sum-of-all-5-digit-numbers-that-can-be-formed-using-0-0-1-1-2-3 .This link is not about non-decreasing order ,but it help you for general types – Not a Salmon Fish Sep 09 '21 at 12:01
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This is a good question First, make 11 dashes and name them according to their position
1 2 3 4 5 6 7 8 9 10 11 So, for example we take a certain value for place 10, let's say it's 1. Then, the possible values for place 11 is 9 i.e. 1,2,3,4,5,6,7,8,9 Similarly if value of place 10 is 2, the possible values for place 11 is 8, and so on. So we can conclude that the total number of values for place 11 if place 1 is fixed = 9+8+7+6+5+4+3+2+1 Now, if we fix place 9, let's say it's value is 1, the total number of combinations for place 10 and place 1 is 45 For 2, it is 45-(the number of combinations of place 11 when place 10 is 1) =45-9=36 For 3, it is 36-(the number of combinations of place 11 when place 10 is 2) =36-8=28 And so on.... Now continue this process till you fix the value of place 1. The equations formed are below:
Place 10 fixed: 9+8+7+6+5+4+3+2+1=45
Place 9 fixed: 45+36+28+21+15+10+6+3+1=165
Place 8 fixed: 165+120+84+56+35+20+10+4+1=495
Place 7 fixed: 495+330+210+126+70+35+15+5+1=1287
Place 6 fixed: 1287+792+462+252+126+56+21+6+1=3003
Place 5 fixed: 3003+1716+924+462+210+84+28+7+1=6435
Place 4 fixed: 6435+3432+1716+782+330+120+36+8+1=12870
Place 3 fixed: 12870+6435+3003+1287+495+165+45+9+1=24310
Place 2 fixed: 24310+11440+5005+2002+715+220+55+10+1=43758 Place 1 fixed: 43758+19448+8008+3003+1001+286+66+11+1=75582 Hence, N=75582
The sum of digits of N=7+5+5+8+2=27
Ans: 27
Hope it helps!!
