Question

Devise an algorithm for generating all the r-combinations of a finite set when repetition is allowed.

Short answer

The algorithm for generating\(r - \)combinations of a finite set when repetition is allowed

\({n_{{c_r}}} = \frac{{n!}}{{r!\left( {n - r} \right)!}}\)

Step-by-step solution

Given

Finite set with repetition is allowed .

\({n_{{c_r}}} = \frac{{n!}}{{r!\left( {n - r} \right)!}}\)

Calculation

The algorithm for generating\(r - \)combinations of a finite set when repetition is allowed if as following procedure next r-combination

\(\left( {{a_1},{a_2},........{a_r}:{\mathop{\rm int}} egers} \right).\)

\(Permutation\{ 1 \le {a_1} \le {a_2} \le {a_3} \le .........{a_r} \le n,{a_1} \ne n\} \)

for\(i = 1\)to\(r\)

While\({b_i} = {a_1}\)

for\(j:1\)to\(r\)

for\(k:1\)to\(n\)

\(\begin{array}{l}{b_j} = {a_k}\\{d_m} = {b_1}{b_2}....{b_r}\end{array}\)

Where \(\{ {d_m}:\) list of combination and\({b_r}\):list of track of element \({d_i}\)}return \({d_{1,}}{d_2}....{d_m} - 1\)

Conclusion:

\({n_{{c_r}}} = \frac{{n!}}{{r!\left( {n - r} \right)!}}\)