Answer:
1. We use permutation given that order is important when it comes to choosing the three from 12 athletes:
P(12, 3)=12!/(12-3)!
P(12, 3)=12!/9!
P(12, 3)=12*11*10*9!/9!
P(12, 3)=12*11*10
P(12, 3)=1320
There are 1320 ways the first, second, and third placers can be chosen.
2.
a. Ways boys can be arranged: 5*4*3*2*1=120
girls : 4*3*2*1=24
boys and girls (treat members of each group all as one entity): 2*1=2 (they can be positioned to the left or right of another)
120*24*12=5760
There are 5760 ways similar members (boys and girls) can stand together.
b. Following the visualization,
B B B B B
G G G G
Ways boys can be arranged: 5*4*3*2*1=120
girls : 4*3*2*1=24
120*24=2880
There are 2880 ways boys and girls can stand alternately.
3. Look for the arrangement of the three who should stay together: 3*2*1=6
Afterwards, treat the three as a single entity when arranged with other participants: 8! or 8*7*6*5*4*3*2*1=40320
6*40320=241920
There are 241920 ways they can be arranged given that three of them should stay together.
4. a. Since they can be arranged in any positions, 5*4*3*2*1=120
b. Children can exchange positions: 3!=6
Both couple can exchange their positions: 2! =2
6*2=12
There are 12 ways of arranging them given this condition.
c. Assuming that parents are the only ones to pose at the front row, similar would be done:
Children can exchange positions: 3!=6
Both couple can exchange their positions: 2! =2
6*2=12
There are 12 ways of arranging them in two rows assuming that only the parents will stay at the front row and their children at the back.
(The result would be changed however if the children were allowed to be in front, which may also combinations.)
Hope this helps!
