Here is a part two of a question, which is homework and I want to make sure of my answer:

A) How many integer numbers with four distinct digits exist that they are either additive or reductive (check my answer)

B)How many four digit integer number exist that that the digits are either Non-decreasing (like 1347,1226,7778) or Non-increasing order (like 6421,6622,9888) ?

My solution for Non-decreasing part : The digits can be repeated so we can construct a four digit number with 4 or 3 or 2 or even one number . By picking 4 numbers out of 9 ( except 0 , because logically it cannot be anywhere in that four digit) there is only one arrangement that matches the property (like 1234)by picking 3 numbers out of 9 there is three arrangements(like 1233,1223,1123) by picking 2.....by picking 1.... So the answer would be like : $$1{9\choose 4}+ 3{9\choose 3}+ 1{9\choose 2}+ 1{9\choose 1}$$

For the Non-increasing part its the same except 0 can be involved as one last or two last or three last ones. So we have : ${9\choose 3}+ {9\choose 2}+ {9\choose 1}$ So the final answer for the increasing part would be :

$$1{9\choose 4}+ 4{9\choose 3}+ 2{9\choose 2}+ 2{9\choose 1}$$

THE FINAL ANSWER FOR PART B is sum of this two answers and because of the OR in the question we have to reduce the common answers in our final answer because we count it twice . The common answers are 1111,222,...,9999 So the final answer is :

$$2{9\choose 4}+ 7{9\choose 3}+ 3{9\choose 2}+ 3{9\choose 1} -9$$

Am I missing somthing or doing something wrong ? I would really appreciate someone check my answer. Thanks in advance.

  • $\begingroup$ Total minus increasing minus decreasing plus constant? $\endgroup$
    – UmbQbify
    Jun 23, 2020 at 13:49
  • $\begingroup$ Also, your calculation is wrong. You have only chosen the digits, they must be arranged as well. $\endgroup$
    – UmbQbify
    Jun 23, 2020 at 13:54
  • 2
    $\begingroup$ @user675453 if i want to construct a four digit number with any four integer(like 1,2,3,4) there is 4! Possibilities but only one of them is in increasing oder (1234) $\endgroup$ Jun 23, 2020 at 13:57
  • $\begingroup$ See the case with 2 digits, these are possible combinations, $(3,1), (2,2) , (1,3)$ here number in ordered pair represents number of times a digit is repeated. $\endgroup$
    – UmbQbify
    Jun 23, 2020 at 13:59
  • $\begingroup$ @user675453 i think you are misunderstanding what the question mean by “Not increasing” and “Not decreasing “ . Numbers like 3294 are not accepted and if we do what you said in the first comment they would be counted.please read the examples of those parts again . $\endgroup$ Jun 23, 2020 at 16:26

1 Answer 1


Let find a correct solution and compare with your numbers.

Obviously a quadruple of numbers can be brought in non-increasing (non-deacreasing) order in a unique way. Therefore it is required only to know how many copies of every digit are present. Essentially it is equivalent to problem of distributing 4 balls among 9 (or 10) bins and can be easily solved by stars and bars method.

If the sequence is non-decreasing it - as you have correctly noted - cannot contain $0$. This means we have choice between $9$ digits, so that the overall count is $$ \binom{4+9-1}4=\binom{12}4=495 (\color{red}{\ne423}).\tag1 $$
If the sequence is non-increasing it can contain up to three $0$. Thus we have choice between $10$ digits, one choice ($0000$) being invalid, so that the count is: $$ \binom{13}4-1=714(\color{red}{\ne552}).\tag2 $$

Together it gives (here you correctly defined the intersection of both sets): $$\binom{12}4+\binom{13}4-10=1200. $$

As seen from (1) and (2) your expressions heavily underestimate the actual numbers.

  • $\begingroup$ Thank you so much . I really appreciate you help . $\endgroup$ Jun 24, 2020 at 5:48
  • $\begingroup$ You are welcome. Let me know if you need help in finding the error in your calculations. $\endgroup$
    – user
    Jun 24, 2020 at 5:51
  • $\begingroup$ I tried to find my error , then I realized in non-increasing part i made a mistake about calculating 0s being involved. My new answer to that part is 219(in a logical way) . When I add 219 (0 being involved ) to 0 not being involved which is 423 as you mentioned , it is equal to 642. And 642 is 72 units less than the correct answer 714 . Also 423 is 72 units less than 495. So my conclusion is that I’m doing the 0 being involved part correct but I’m missing somthing about 0 not being involved in non-increasing part . $\endgroup$ Jun 24, 2020 at 6:57
  • $\begingroup$ and ${\binom 9 2}=72$ so i think I’m missing somthing in calculating possible ways of constructing a four digit number with 2 integers (when 0 is not invloved). $\endgroup$ Jun 24, 2020 at 7:00
  • $\begingroup$ You are correct. The error is there. The factor should be 3 instead of 1. $\endgroup$
    – user
    Jun 24, 2020 at 7:04

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