Question

A total of 6 family members will appear in a portrait. If 2 of the family members may sit in the front row, how many combinations of family members may sit in the front row?

Answer 1

15

Explanation:

We're asked specifically about combinations, which means that we don't care in what order the two people sit in the front row (Mom on left and Dad on right is the same as Dad on left and Mom on right).

The general formula for combinations is

`C_(n,k)=(n!)/((k!)(n-k)!)` with `n="population", k="picks"`

`C_(6,2)=(6!)/((2!)(6-2)!)=(6!)/(2!4!)=(6xx5xx4!)/(2xx4!)=15`

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As an aside, if it did matter who sat where in the front row, we'd use a permutation:

`P_(n,k)=(n!)/((n-k)!); n="population", k="picks"`

`P_(6,2)=(6!)/((6-2)!)=(6!)/(4!)=30`