Count restricted selections and calculate exact probabilities

Group labels do not make individual objects identical

If six labelled red tokens and four labelled blue tokens are available, two selections with different red token identities are different, even when their colour composition is the same. Use combinations within each category, then multiply across categories

From six distinct red tokens and four distinct blue tokens, select exactly two red and one blue. How many unordered selections are possible?

Required and excluded members change the pool

For a group of k from n distinct objects with one specified object required, choose the remaining k−1 from n−1. With one specified object excluded, choose all k from n−1. These constraints refer to one individual, not every member of a colour category

Select four people from nine, with one specified person required. How many groups exist?

At least one is often easiest by complement

For three objects selected from six distinct red and four distinct blue objects, count all C(10,3)=120 selections and subtract the C(4,3)=4 with no red. This gives 116 with at least one red. The complement is taken inside the same fixed-size selection space

Derive a two-group identity, ungraded

For disjoint sets of m and n distinct objects, prove the sum over j of C(m,j)C(n,k−j) equals C(m+n,k). Explain why every k-element selection occurs in exactly one summand. Use coefficient zero outside its valid range

Model: partition by composition

Count all k-element subsets of the union directly as C(m+n,k). Alternatively classify a subset by its uniquely determined number j from the first group. There are C(m,j) choices from that group and C(n,k−j) from the second. These composition classes are disjoint and exhaustive, so their sum equals the direct count. Impossible composition classes contribute zero, including when j or k−j is outside the group size