2 Ways to do this:

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Because the parents are the limiting factor here - at least one must be in the final group - it can be helpful to begin by thinking through the situations that will work: Mom goes and Dad doesn't; Dad goes and Mom doesn't; and both parents go (this is the one people forget!). With each of those situations:

Mom goes, Dad doesn't: that leaves 3 spots left to select from the 7 children. Using the combinations formula that gives 7!/3!4!=7∗6∗5/3∗2=35

Dad goes, Mom doesn't: that math is the same as above, with 7 children vying for the 3 remaining spots. Using the combinations formula that's: 7!/3!4!=7∗6∗5/3∗2=35

Both parents go: that leaves the 7 children with only 2 spots to be chosen. The combinations formula then looks like: 7!/2!5!=7∗6/2=21

So the three different possible situations have 35, 35, and 21 outcomes each. The sum of those is 91 total options that will work.

**Quote:**

Alternatively, you could start with the total number of unrestricted groups (9 family members, choose 4) and work backward by subtracting those that violate the constraint (those groups that consist of 4 children with no parents). Were there no restriction and there were 9 people vying for 4 spots, that math would be:

9!/4!5!=9∗8∗7∗6/4∗3∗2∗1=126

But then you must subtract the "no parents / all children" groups. To find those, take the parents out of the pool of candidates, meaning that there are 7 people from which you'll select 4. That math is 7!/3!4!=7∗6∗5/3∗2=35

Therefore, 126 - 35 = 91, again leading to answer choice C.

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