The other way to solve this involves no math at all: The question gives you three pieces of data, presumably all important to solving the problem (16 gifts, 4 children, 4 gifts for each child). The only answer that includes all three pieces of data in it is C. Sometimes multiple choice questions are a matter of logic and not knowing the math.

Interesting… but I don’t think that method is necessarily valid and applies to all cases. For instance, if I removed the constraint for 4 gifts per child, then I would have a far more difficult problem, not, as your logic would suggest, an easier problem with fewer moving parts, so to speak.

At the same time, I think often you can work through a problem based on logic. However, with this problem I think your assumption, though it got you the right answer, would not pertain in other cases.

The answer is C.
Firstly for selecting 4 gifts for a child, the number of possible combinations is 16C4.
For the next child, the possible selections are 12C4 and for the third, it’s 8C4.
On multiplying all the possible selections, the value obtained is 16/(4!)^4.
That gives the final solutions.
Also, as the arrangement of the gifts is not important, the answer cannot be 16!/4!.

Clemmonsdogpark blog comment policy: To create the best experience for our readers, we will only approve comments that are relevant to the article, general enough to be helpful to other students, concise, and well-written! 😄 Due to the high volume of comments across all of our blogs, we cannot promise that all comments will receive responses from our instructors.

We highly encourage students to help each other out and respond to other students' comments if you can!

If you are a Premium Clemmonsdogpark student and would like more personalized service from our instructors, you can use the Help tab on the Clemmonsdogpark dashboard. Thanks!

The other way to solve this involves no math at all: The question gives you three pieces of data, presumably all important to solving the problem (16 gifts, 4 children, 4 gifts for each child). The only answer that includes all three pieces of data in it is C. Sometimes multiple choice questions are a matter of logic and not knowing the math.

Interesting… but I don’t think that method is necessarily valid and applies to all cases. For instance, if I removed the constraint for 4 gifts per child, then I would have a far more difficult problem, not, as your logic would suggest, an easier problem with fewer moving parts, so to speak.

At the same time, I think often you can work through a problem based on logic. However, with this problem I think your assumption, though it got you the right answer, would not pertain in other cases.

The answer is C.

Firstly for selecting 4 gifts for a child, the number of possible combinations is 16C4.

For the next child, the possible selections are 12C4 and for the third, it’s 8C4.

On multiplying all the possible selections, the value obtained is 16/(4!)^4.

That gives the final solutions.

Also, as the arrangement of the gifts is not important, the answer cannot be 16!/4!.