Finding the ways to arrange 6 out of 10 people. 2 of them must be together I have a group of 10 people and I need to select 6 of them to pose for a photograph. Two of the 10 people are named Ken and Karen. How many ways can I arrange 6 people out of the group of 10 for the photograph if a) Ken and Karen must be in the photograph? b) Ken and Karen must be in the photograph and they must appear next to each other in the photograph For part a), Since only 6 out of 10 can be selected to pose for a photograph, It will be 10C6. Within that 6 people, there must be Ken and Karen. 6C2 Hence, I multiplied them together 10C6 * 6C2 = 3150. The correct answer is 50400. Why is that so? Can someone correct my logic, please. Since I got part a) wrong, I couldn't start on part b)

> How many ways can six people out of the group of $10$ be arranged for the photograph if Ken and Karen must be in the photograph.

Since Ken and Karen are in the photograph, we must select four of the other eight people. The six selected people can then be arranged in $6!$ ways. Hence, the number of permissible arrangements is $$\binom{8}{4}6!$$