MHB Solving Permutations Question with Restrictions

[Lancelot1]

Hello all

Please look at this questions:

What is the number of permutations for creating a code of 3 digits from the digits 1,2,3,...,9 , such that every digit is equal or larger from the previous one ?

I know that if I wanted the number of permutations without restrictions it would be:

\[9^{3}\]

If all digits can be the same. Otherwise it would be

\[9\cdot 8\cdot 7\]

How do I approach the restriction, I mean, there are many possibilities, it depends on the numbers that I get.

Thank you.

 

[Evgeny.Makarov]

The number of ordered tuples $(a,b,c)$ where $1\le a\le b\le c\le 9$ is equal to the number of multisets $\{a,b,c\}$ because every multiset can be ordered in ascending order in a unique way.

 

[Lancelot1]

The number of ordered tuples $(a,b,c)$ where $1\le a\le b\le c\le 9$ is equal to the number of multisets $\{a,b,c\}$ because every multiset can be ordered in ascending order in a unique way.

Do you mean that the answer is :

\[\binom{11}{3}\] ?

I don't understand the rational behind it.

 

[Evgeny.Makarov]

Do you mean that the answer is :

\[\binom{11}{3}\] ?

Yes. Because there is a bijection $f$ between the set of multisets of size 3 and the set of non-decreasing triples. Namely, $f(s)$ is $s$ sorted in increasing order.