A total of 6 family members will appear in a portrait. If 3 of the family members may sit in the front row, how many combinations of family members may sit in the front row?
Let's say the front row has 3 slots: A,B,C there are 6 choices available to fill slot A then 5 choices for slot B and 4 choices for slot C Multiply these choices out to get 6*5*4 = 120 This would be the answer if order mattered. But it doesn't. Since order does NOT matter, we divide by 3! = 3*2*1 = 6. Why? Because for each grouping, there are 6 ways to order the 3 people. Example: ABC is the same as BAC or CBA So 120/6 = 20 is the true answer. There are 20 combinations here. Notice how \(\LARGE _{6}C_{3} = 20\). This is no coincidence. You can use the combination formula to get the same answer.
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