MHB Solving Permutations Question with Restrictions

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The discussion focuses on calculating the number of permutations for a 3-digit code using the digits 1 to 9, with the restriction that each digit must be equal to or larger than the previous one. The initial approach considers unrestricted permutations, leading to calculations of \(9^{3}\) for identical digits and \(9 \cdot 8 \cdot 7\) for distinct digits. The key insight is that the problem can be transformed into counting multisets, where the number of ordered tuples \((a,b,c)\) corresponds to the number of multisets of size 3. The final answer is determined to be \(\binom{11}{3}\), supported by a bijection between multisets and non-decreasing triples. This method effectively resolves the permutation question under the given restrictions.
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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.
 
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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.
 
Evgeny.Makarov said:
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.
 
Lancelot said:
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.
 

Thread 'Area of Overlapping Squares'

Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?
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