Task

Given non-negative integers    m    and    n,   generate all size    m    combinations   of the integers from    0   (zero)   to    n-1    in sorted order   (each combination is sorted and the entire table is sorted).



Example

3   comb    5      is:

0 1 2
0 1 3
0 1 4
0 2 3
0 2 4
0 3 4
1 2 3
1 2 4
1 3 4
2 3 4

If it is more "natural" in your language to start counting from    1   (unity) instead of    0   (zero),

the combinations can be of the integers from    1   to    n. 



See also


 The number of samples of size k from n objects.


With   combinations and permutations   generation tasks. 






 Order Unimportant

 Order Important


 Without replacement








(


n
k


)




=

n



C

k


=



n
(
n
−
1
)
…
(
n
−
k
+
1
)


k
(
k
−
1
)
…
1





{\displaystyle {\binom {n}{k}}=^{n}\operatorname {C} _{k}={\frac {n(n-1)\ldots (n-k+1)}{k(k-1)\dots 1}}}










n



P

k


=
n
⋅
(
n
−
1
)
⋅
(
n
−
2
)
⋯
(
n
−
k
+
1
)


{\displaystyle ^{n}\operatorname {P} _{k}=n\cdot (n-1)\cdot (n-2)\cdots (n-k+1)}




Task: Combinations

Task: Permutations


 With replacement








(



n
+
k
−
1

k


)




=

n
+
k
−
1



C

k


=



(
n
+
k
−
1
)
!


(
n
−
1
)
!
k
!





{\displaystyle {\binom {n+k-1}{k}}=^{n+k-1}\operatorname {C} _{k}={(n+k-1)! \over (n-1)!k!}}








n

k




{\displaystyle n^{k}}




Task: Combinations with repetitions

Task: Permutations with repetitions



next_comb(n, p, a) := block(
   [a: copylist(a), i: p],
   if a[1] + p = n + 1 then return(und),
   while a[i] - i >= n - p do i: i - 1,
   a[i]: a[i] + 1,
   for j from i + 1 thru p do a[j]: a[j - 1] + 1,
   a
)$
 
combinations(n, p) := block(
   [a: makelist(i, i, 1, p), v: [ ]],
   while a # 'und do (v: endcons(a, v), a: next_comb(n, p, a)),
   v
)$
 
combinations(5, 3);
/* [[1, 2, 3], 
    [1, 2, 4], 
    [1, 2, 5], 
    [1, 3, 4], 
    [1, 3, 5], 
    [1, 4, 5], 
    [2, 3, 4], 
    [2, 3, 5], 
    [2, 4, 5], 
    [3, 4, 5]] */