The idea is, can we derive combination formula using recursion?
Let us start with a particular case.
How many ways can we make groups of 2 out of n people?
Let T(n) be the answer for n people.
Let us think wishfully and see if we can get to solution for n when solution for n-1 is known.
[n] | [# of pairs] | [pairs] |
2 - A,B | 1 | (A,B) |
3 - A,B,C | 3 | (A,B);(B,C);(C,A) |
4 - A,B,C,D | 6 | (A,B);(B,C);(C,D);(D,A);(A,C);(B,D) |
From above table, it is easy to observe that
T(n) = T(n-1) + (n-1)
We can solve this recurrence using substitution and see that
T(n) = n(n-1)/2
which is indeed (n combination 2).
How many ways can we make group of k ppl out of n ppl?
Let it be denoted by T(n,k).
Clearly, T(k,k) = 1
We can apply the process as in above case by solving for T(n,3), T(n,4) etc and easily see that
T(n,k) = T(n-1, k) + T(n-1, k-1)
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