1)

Year Height
0 4
1 4+ x
2 4 + 2x
3 4 + 3x
4 4 + 4x
5 4 + 5x
6 4 + 6x

(4 + 4x)1.2 = 4 + 6x
4.8+ 4.8x = 4 + 6x
.8 = 1.2x
x = 2/3


2)

We need to multiply the total number of options for chairs by the total number of options for tables.

Chairs = 5C2 = 5!/3!2! = 10
This means that the total combinations of tables must be 15. (150/10 = 15)

Clue here is if 5C2 = 10, we must be choosing from more options. 6 ends up being the answer. I think of these problems in "spots." There are 2 spots available so multiply the number of options you have: x(x-1). Then you divide by #spots factorial, or 2! in this case. So we are looking for two consecutive numbers that = 30 (15 x 2).


3) If 6 machines can do a job in 12 days, then 1 machine will take 72 days (6 x 12). Now we can divide 72 by 8 days to get 9 machines. This 3 more than 6.

Thank you! I am a bit confused on the second one. Here is another example that may help you explain where I am lost. I might have the concept wrong.


Quality control check 2 lightbulbs from a box of 12. How many different samples can be choosen?
11*6=66

When I approach this problem I think I get thrown off when picking more than one items in a group. For instance if you have three shirts, two pairs of shorts, and two pairs of shoes you can have 12 different outfits to choose from.