Let's say the children are named A, B, C, and D

Stage 1: Select 4 gifts to give to child A
Since the order in which we select the 4 gifts does not matter, we can use combinations.
We can select 4 gifts from 16 gifts in 16C4 ways (= 16!/(4!)(12!))
So, we can complete stage 1 in 16!/(4!)(12!) ways

Stage 2: select 4 gifts to give to child B
There are now 12 gifts remaining
Since the order in which we select the 4 gifts does not matter, we can use combinations.
We can select 4 gifts from 12 gifts in 12C4 ways (= 12!/(4!)(8!))
So, we can complete stage 2 in 12!/(4!)(8!) ways


Stage 3: select 4 gifts to give to child C
There are now 8 gifts remaining
We can select 4 gifts from 8 gifts in 8C4 ways (= 8!/(4!)(4!))
So, we can complete stage 3 in 8!/(4!)(4!) ways

Stage 4: select 4 gifts to give to child D
There are now 4 gifts remaining
NOTE: There's only 1 way to select 4 gifts from 4 gifts, but if we want the answer to look like the official answer, let's do the following:
We can select 4 gifts from 4 gifts in 4C4 ways (= 4!/4!)
So, we can complete stage 4 in 4!/4! ways

By the Fundamental Counting Principle (FCP), we can complete all 4 stages (and thus distribute all 16 gifts) in [16!/(4!)(12!)][12!/(4!)(8!)][8!/(4!)(4!)][4!/4!] ways

A BUNCH of terms cancel out to give us = 16!/(4!)⁴

Answer: C

Note: the FCP can be used to solve the MAJORITY of counting questions on the GRE. So, be sure to learn it.

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