A family consisting of 2 parents and 3 children is to pose a picture with 2 family members in the front and 3 in the back.

Question

a)How many arrangements are possible with no restrictions?
b) How many arrangements are possible if the parents must sit in the front? c) How many arrangements are possible if parents must be next to each other?

My answers are a) 5! b) 3! for arrangements of children in the back and times two for two arrangements of parents 3! x 2! c) 2! x 3! x 4 (4 is the number of places where parents can be)

jvdhooft · Answer 1 · 2017-06-15T08:39:48.853

5

Your answers to a) and b) are correct, but your answer to c) is wrong. If the parents must be next to each other, there are only three possible arrangements with the parents $P$ and the children $C$:

$$PPC|CC, CPP|CC, CCC|PP$$

As such, the number of valid arrangements equals:

$$3 \cdot 2! \cdot 3! = 36$$

edited Jun 15 '17 at 08:39

answered Jun 15 '17 at 08:36

jvdhooft

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$3\times 2!\times 3!=36$ – CY Aries Jun 15 '17 at 08:39

score 4 · Accepted Answer · answered Jun 15 '17 at 08:45

Your C) is wrong.

If the parents are in the front, there are $(2!)(3!)=12$ ways to order the parents and the kid.

If the parents are in the back, then there are $(2!)(2)(3)(2!)=24$ to order the parents, choose the left or the right block to be, to choose one of the kids to be in the back with them, and to order the kids in the front.

Finally, we have $12+24=\boxed{36}$ ways.

A family consisting of 2 parents and 3 children is to pose a picture with 2 family members in the front and 3 in the back.

2 Answers2