Yes, each would have 14, as 14 x 4 = 56.
Carefully arrange 12 rows with 8 coins in each row.
Two ways only. 4 rows with 25 stamps each or 5 rows with 20 stamps each.
90000
wait not the discription srry
You, can't.
You can't arrange 12 apples into 5 rows with four in each. To do that you would need 20 apples because 5 rows of 4 = 20 or 5x4=20. You can arrange 12 apples into 3 groups of 4 because 3x4=12
4
Carefully arrange 12 rows with 8 coins in each row.
You can have: 1 row of 36 2 rows of 18 3 rows of 12 4 rows of 9 or 6 rows of 6, so in total there are 5 ways.
There are eight possible combinations... 1 row of 24 cans 2 rows of 12 cans 3 rows 8 cans 4 rows of 6 cans 6 rows of 4 cans 8 rows of 3 cans 12 rows of 2 cans 24 rows of 1 can
Two ways only. 4 rows with 25 stamps each or 5 rows with 20 stamps each.
in the form of a start
oooo arrange them in a triangle shape, four coins each side. o o oo o
You can show 24 cans in one row, 12 cans in 2 rows, 8 cans in 3 rows, and 6 cans in 4 rows.
Four ways:1 row with 6 in each row.2 rows with 3 in each row.3 rows with 2 in each row.6 rows with 1 in each row.
2 rows of 18 squares3 rows of 12 squares4 rows of 9 squares6 rows of 6 squares9 rows of 4 squares12 rows of 3 squares18 rows of 2 squares36 rows of 1 squareI would not count "1 row of 36 squares", because you only have a single row that cannot equal another row (there is only one rowafter all). If this is for homework, I would state your reasoning for excluding (or including) that set. Count all the options up, and you have 8 different ways you can arrange the rows with the exclusion.
by 2 rows