Given you have 6 questions.
Each question can have marks between 0 to 10
The following question must be less or equal to the previous question.
How many combination of set of marks is possible?
First time posting and this is my solution, hope it right.
Case 1: 6 questions have different number of marks.
combination of set will be (9*8*7*6*5*4)/(6!)
Explain: the first question has 9 choices of number of marks. the second one has 8 choices. the third one has 7 choices. the fourth one has 6 choices. the fifth one has 5 choices and the last one has 4 choices. and with condition that the following question must be less or equal to the previous question, so we have to divided by 6! because with 6 different numbers (excluded 0), we have 6! cases to put them together but there is only one case that suit the condition.
Case 2: there are 2 questions have the same number of marks.
let say combine these 2 questions to one question. so we have five questions now.
therefore, the combination of set in this case will be (9*8*7*6*5)/(5!)
Case 3: 3 questions have the same number of marks. combination of set: (9*8*7*6)/(4!)
Case 4: 4 questions have the same number of marks. combination of set : (9*8*7)/(3!)
case 5: 5 questions have the same number of marks. combination of set : (9*8)/(2!)
case 5: 6 questions have the same number of marks. combination of set : 9
by adding up all 6 case together we will get the solution: 465
I will explain it to you using an equivalent example which is easy to understand.
Consider the first eleven letters of the English alphabet: [TEX]\{A,B,C,D,E,F,G,H,I,J,K\}/TEX].
Select any multi-set of six. (multi-sets allow repeating members)
Say we select $\displaystyle [E,K,K,A,I,I]$
There are $\displaystyle \dbinom{16}{6}$ such possible selections.
This string can be rearranged in $\displaystyle \dfrac{6!}{2!\cdot2!}$ ways.
BUT only of those arrangements is in alphabetical order.
If the question were "how many six letter strings using the first eleven letters( repeats allowed) have there letters in alphabetical order?" The answer is $\displaystyle \dbinom{16}{6}$,
That is the same as the original question.