I got this far in my working out and am already stuck. have i gone wrong?
1 answer:
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Answer:
1) 1 element
2) 13 elements
3) 22 elements
4) 40 elements
Step-by-step explanation:
1) Only one element will have no tails: the event that all the coins are heads.
2) 13 elements will have exactly one tile. Basically you have one element in each position that you can put a tail in.
3) There are elements that have exactly 2 tails. From those elements we have to remove the only element that starts and ends with a tail and in the middle it has heads only and the elements that starts and ends with a head and in the 11 remaining coins there are exactly 2 tails. For the last case, there are
possibilities, thus, the total amount of elements with one tile in the border and another one in the middle is 78-55-1 = 22
4) We can have:
- A pair at the start/end and another tail in the middle (this includes a triple at the start/end)
- One tail at the start/end and a pair in the middle (with heads next to the tail at the start/end)
For the first possibility there are 2 * 11 = 22 possibilities (first decide if the pair starts or ends and then select the remaining tail)
For the second possibility, we have 2*9 = 18 possibilities (first, select if there is a tail at the end or at the start, then put a head next to it and on the other extreme, for the remaining 10 coins, there are 9 possibilities to select 2 cosecutive ones to be tails).
This gives us a total of 18+22 = 40 possibilities.
Please refer to the diagrams to understand these answers:
CUBE (1):
Paint on cube (1) contains 3.125% of the paint.
CUBE (2):
Paint on cube (2) contains 2.083% of the paint.
CUBE (3):
Paint on cube (3) contains 1.042% of the paint.
CUBE (4):
Contains 0% of the paint.
-----------------
There would be four different types of cubes which would contain different amounts of paint.
Some cubes in fact would have no paint (cube (4)). Now we can check if our results are correct by counting how many cube (1)'s, cube (2)'s, cube (3)'s and cube (4)'s there are...
Cube (1) x 8 + Cube (2) x 24 + cube (3) x 24 + cube (4) x 8
= 3.125% x 8 + 2.083% x 24 + 1.042% x 24 + 0% x 8
= 1
= 100%
CUBE (1):
Paint on cube (1) contains 3.125% of the paint.
CUBE (2):
Paint on cube (2) contains 2.083% of the paint.
CUBE (3):
Paint on cube (3) contains 1.042% of the paint.
CUBE (4):
Contains 0% of the paint.
-----------------
There would be four different types of cubes which would contain different amounts of paint.
Some cubes in fact would have no paint (cube (4)). Now we can check if our results are correct by counting how many cube (1)'s, cube (2)'s, cube (3)'s and cube (4)'s there are...
Cube (1) x 8 + Cube (2) x 24 + cube (3) x 24 + cube (4) x 8
= 3.125% x 8 + 2.083% x 24 + 1.042% x 24 + 0% x 8
= 1
= 100%