Partial dichotomy problem

Five volunteers were divided into three groups, 2 in each group, and the other group 1.

[C5(2)C3(2)c 1( 1)]/[A2(2)]

= 15

Then they will go to three different venues of the World Expo to provide services, and the different allocation scheme is A3(3).

=6

So, five volunteers were divided into three groups, two in each group, and the other group 1 children.

There are three different venue services for the World Expo, and the different allocation schemes are as follows.

15*6=90 species