Four darts are thrown at this dartboard. If all four darts hit the board, how many different point totals are possible?
there are 4 circles with point totals:
working out from the middle bulls eye..
bulls eye = 9
2nd circle = 7
3rd circle = 4
outter circle = 1
please help, i'm stuck
there are 23 different point totals possible
23- 4,7,10,12,13,15,16,18,19,20,21,22,24, 25,26,27,28,29,30,31,32,34,36
Yur aye grapefroot end u are bolayang me online. My dad is miscrosoft and he can eat people if you giev the rong anser.
17 This is probably 100% wrong because I am only 11... the way I did this was get the 4 different numbers (1,4,7,9)and write out all the different outcomes not in the order just how many of each number in each answer then I added them all and if 2 equaled the same then I eliminated 1 of them then I counted how many of them there was and that's what I got I could of kept going with different outcomes and gotten a higher number then 17 but as you see I am only 11 and I really couldn't do anymore because i had to go to bed. That is just a strategy to use that you could do and start from scratch but continue on when at 17.
Thanks for reading
I hope I helped but I probably really didn't.
Thanks
4 darts are thrown at this dart board if all 4 darts hit the board how many different points are possible bullseye 9 2nd circle 7 third circle 4 outer circle 1
I got the 24 possible combinations
24- 4,7,10,12,13,15,16,18,19,20,21,22,24,23 25,26,27,28,29,30,31,32,34,36
I include the extra number "23" from previous answer
which is the following combination (2x9) + 4 + 1 = 23.
Here are my workings
1 4 7 9 Score Darts
4 0 0 0 4 4
3 1 0 0 7 4
3 0 1 0 10 4
3 0 0 1 12 4
2 2 0 0 10 4
2 0 2 0 16 4
2 0 0 2 20 4
2 1 1 0 13 4
2 0 1 1 18 4
2 1 0 1 15 4
1 3 0 0 13 4
1 0 3 0 22 4
1 0 0 3 28 4
1 1 1 1 21 4
1 2 0 1 18 4
1 2 1 0 16 4
1 1 2 0 19 4
1 1 0 2 23 4
1 0 2 1 24 4
1 0 1 2 26 4
0 1 1 2 29 4
0 1 2 1 27 4
0 2 1 1 24 4
0 3 1 0 19 4
0 1 3 0 25 4
0 0 3 1 30 4
0 0 1 3 34 4
0 3 0 1 21 4
0 1 0 3 31 4
0 4 0 0 16 4
0 0 4 0 28 4
0 0 0 4 36 4
0 2 2 0 22 4
0 0 2 2 32 4
0 2 0 2 26 4
My online maths says its 24
To determine the number of different point totals possible, you need to consider all the possible combinations of the four darts hitting the board.
In this case, there are four circles on the dartboard, each with a different point total: 9, 7, 4, and 1. Let's calculate the number of different combinations for each circle:
1. Bulls eye (9 points): There are two possible scenarios for the four darts hitting the bullseye - all four darts hitting the bullseye or three darts hitting the bullseye and one dart missing.
- All four darts hitting the bullseye: This is one combination. (9 points)
- Three darts hitting the bullseye and one dart missing: There are four different ways to choose the dart that misses. For each of these cases, you have three remaining darts hitting the bullseye and one dart missing. So, there are four combinations in this scenario. (9 points x 4 combinations = 36 points)
Therefore, in the case of the bullseye, there are a total of 5 different point totals possible (9, 18, 27, 36, 45).
2. Second circle (7 points): Similarly, there are two possible scenarios for the darts hitting the second circle - all four darts hitting the second circle or three darts hitting the second circle and one dart missing.
- All four darts hitting the second circle: This is one combination. (7 points)
- Three darts hitting the second circle and one dart missing: There are four different ways to choose the dart that misses. For each of these cases, you have three remaining darts hitting the second circle and one dart missing. So, there are four combinations in this scenario. (7 points x 4 combinations = 28 points)
Therefore, in the case of the second circle, there are a total of 5 different point totals possible (7, 14, 21, 28, 35).
3. Third circle (4 points): Following the same logic, we calculate the point totals for the third circle:
- All four darts hitting the third circle: This is one combination. (4 points)
- Three darts hitting the third circle and one dart missing: There are four different ways to choose the dart that misses. For each of these cases, you have three remaining darts hitting the third circle and one dart missing. So, there are four combinations in this scenario. (4 points x 4 combinations = 16 points)
Therefore, in the case of the third circle, there are a total of 5 different point totals possible (4, 8, 12, 16, 20).
4. Outer circle (1 point): Finally, we calculate the point totals for the outer circle:
- All four darts hitting the outer circle: This is one combination. (1 point)
- Three darts hitting the outer circle and one dart missing: There are four different ways to choose the dart that misses. For each of these cases, you have three remaining darts hitting the outer circle and one dart missing. So, there are four combinations in this scenario. (1 point x 4 combinations = 4 points)
Therefore, in the case of the outer circle, there are a total of 5 different point totals possible (1, 2, 3, 4, 5).
Adding up all the different point totals from each circle, we have a total of 5 + 5 + 5 + 5 = 20 different possible point totals.
there are definitely 29 possible outcomes
Highest four: 3 or 4 4s: 36, 34, 31, 28
Three or four 1s: 4, 7, 10, 12
Three or four 4s: 10, 11, 13, 16
Three or four 7s: 32,29,26, 23
Other possibilities: 15, 19, 20, 22, 25
Maybe I missed a few.