In how many ways can 8 different gifts be divided among four children such that each child receives exactly two gifts?  Choose the correct choice from the following choices
A) 16^4
B) (4!)^4
C) 8!/(2!)^4
D) 8!/4!
E) 4^16
Total 8 different Gifts, and 4 children.   Thus any one child gets 8C2 gifts,   then the other child gets 6C2 gifts(8 total - 2 already given),   then the third one gets 4C2 gifts,   and the last child gets 2C2 gifts.   Since order in which each child gets the gift is not imp, thus, ans :   88!/(2!)^42 * 68!/(2!)^42 * 48!/(2!)^42 * 28!/(2!)^42= 8! / (2!)^4   Ans : 8!/(2!)^4.
<result>C</result>
In how many ways can 8 different gifts be divided among four children such that each child receives exactly two gifts?