Mathematical permutation and combination problem? It is not difficult to solve a problem. .

Five volunteers were assigned to three venues, and each venue required at least one person.

Then the number of people allocated to each venue can only be (1, 1, 3) and (1, 2,2).

Consider the combination of (1, 1, 3).

There are c (5, 1) ways to choose 1 person from 5 people first.

There are C (4, 1) ways to choose 1 from the remaining four people.

Finally, there is the method of C (3,3) to select 3 people from the remaining 3 people.

Therefore, for the combination of (1, 1, 3),

There are c (5, 1) * c (4,1) * c (3,3) = 20 methods;

Similarly, for the combination of (1, 2,2),

There are C(5,1) * c (4,2) * c (2,2) = 30 methods.

Because,

(1, 1, 3) combination If the arrangement order is considered,

There are three possibilities (1, 1, 3) (1, 3, 1) (3,1);

(1, 2, 2) combination If the arrangement order is considered,

There are three possibilities (1, 2,2) (2, 1, 2) (2,2,1);

So,

The final result is: 20 * 3+30 * 3 = 150 methods.