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Since we can map such sequence

$$0\leq a_1\leq a_2\leq a_3 \leq \cdots \leq a_{n-1}\leq a_n < m $$ to $$0 < b1=a_1+1 < b2=a_2+2 < b3=a_3+3 <\cdots < b_n=a_n+n < m+n $$ and $\{b_1,b_2,\cdots b_n\}$ is the n subsets of Range[m+n-1]

And we can get $\{a_1,a_2,\cdots a_n\}$ from $\{b_1,b_2,\cdots b_n\}-\{1,2,\cdots,n\}$

m = 8;
n = 5;
list = Subsets[Range[m+n-1], {n}]
Subtract[#, Range[n]] & /@ list

Since we can map such sequence

$$0\leq a_1\leq a_2\leq a_3 \leq \cdots \leq a_{n-1}\leq a_n < m $$ to $$0 < b_1 = a_1+1 < b_2 = a_2+2 < b_3 =a_3+3 <\cdots < b_n=a_n+n < m+n $$ and $\{b_1,b_2,\cdots b_n\}$ is the n subsets of Range[m+n-1]

And we can get $\{a_1,a_2,\cdots a_n\}$ from $\{b_1,b_2,\cdots b_n\}-\{1,2,\cdots,n\}$

m = 8;
n = 5;
list = Subsets[Range[m+n-1], {n}]
Subtract[#, Range[n]] & /@ list

Since we can map such sequence

$$0\leq a_1\leq a_2\leq a_3 \leq \cdots \leq a_{n-1}\leq a_n < m $$ to $$0 < b1=a_1+1 < b2=a_2+2 < b3=a_3+3 <\cdots < b_n=a_n+n < m+n $$ and $\{b_1,b_2,\cdots b_n\}$ is the n subsets of Range[m+n]

And we can get $\{a_1,a_2,\cdots a_n\}$ from $\{b_1,b_2,\cdots b_n\}-\{1,2,\cdots,n\}$

m = 8;
n = 5;
list = Subsets[Range[m], {n}]
Subtract[#, Range[n]] & /@ list

Since we can map such sequence

$$0\leq a_1\leq a_2\leq a_3 \leq \cdots \leq a_{n-1}\leq a_n < m $$ to $$0 < b1=a_1+1 < b2=a_2+2 < b3=a_3+3 <\cdots < b_n=a_n+n < m+n $$ and $\{b_1,b_2,\cdots b_n\}$ is the n subsets of Range[m+n-1]

And we can get $\{a_1,a_2,\cdots a_n\}$ from $\{b_1,b_2,\cdots b_n\}-\{1,2,\cdots,n\}$

m = 8;
n = 5;
list = Subsets[Range[m+n-1], {n}]
Subtract[#, Range[n]] & /@ list

Since we can map such sequence

$$0\leq a_1\leq a_2\leq a_3 \leq \cdots \leq a_{n-1}\leq a_n < m $$ to $$0 < b1=a_1+1 < b2=a_2+2 < b3=a_3+3 <\cdots < b_n=a_n+n < m+n $$ and $\{b_1,b_2,\cdots b_n\}$ is the n subsets of Range[m+n]

And we can get $\{a_1,a_2,\cdots a_n\}$ from $\{b_1,b_2,\cdots b_n\}-\{1,2,\cdots,n\}$

list = With[{m = 10, n = 5}, Subsets[Range[m + n], {n}]]

Since we can map such sequence

$$0\leq a_1\leq a_2\leq a_3 \leq \cdots \leq a_{n-1}\leq a_n < m $$ to $$0 < b1=a_1+1 < b2=a_2+2 < b3=a_3+3 <\cdots < b_n=a_n+n < m+n $$ and $\{b_1,b_2,\cdots b_n\}$ is the n subsets of Range[m+n]

And we can get $\{a_1,a_2,\cdots a_n\}$ from $\{b_1,b_2,\cdots b_n\}-\{1,2,\cdots,n\}$

m = 8;
n = 5;
list = Subsets[Range[m], {n}]
Subtract[#, Range[n]] & /@ list