I have 4 arrows. I hit a target with all four shots. With each shot I could have scored 25, 10, 5, 1 points.?

Question

How many total scores are possible?

Answer 1

Well, we know the answer is no more than 100. :) In fact, it's no more than 97.

There's no really short way to do it. I suggest that you work it out separately for the 5 cases in which there are 0, 1, 2, 3, or 4 25s, and then add up those results.

Answer 2

Possible scores:
4*25 = 100
3*25, 1*10 = 85
3*25, 1*5 = 80
3*25, 1*1 = 76
2*25, 2*10 = 70
2*25, 1*10, 1*5 = 65
2*25, 1*10, 1*1 = 61
2*25, 2*5 = 60
2*25, 1*5, 1*1 = 56
2*25, 2*1 = 52
1*25, 3*10 = 55
1*25, 2*10, 1*5 = 50
1*25, 2*10, 1*1 = 46
1*25, 1*10, 2*5 = 45
1*25, 1*10, 1*5, 1*1 = 41
1*25, 1*10, 2*1 = 37
1*25, 3*5 = 40
1*25, 2*5, 1*1 = 36
1*25, 1*5, 2*1 = 32
1*25, 3*1 = 28
4*10 = 40 (already got this)
3*10, 1*5 = 35
3*10, 1*1 = 31
2*10, 2*5 = 30
2*10, 1*5, 1*1 = 26
2*10, 2*1 = 22
1*10, 3*5 = 25
1*10, 2*5, 1*1 = 21
1*10, 1*5, 2*1 = 17
1*10, 3*1 = 13
4*5 = 20
3*5, 1*1 = 16
2*5, 2*1 = 12
1*5, 3*1 = 8
4*1 = 4

So there are 34 possible total scores.

Note that a quick upper bound is
(all four the same): 4.1 = 4
+ (3 one value, 1 another value): 4.3 = 12
+ (2, 2): 4.3 / 2! = 6
+ (2, 1, 1): 4.3.2 / 2! = 12
+ (1, 1, 1, 1): 4.3.2.1 / 4! = 1
for a total of 35. But this doesn't find cases where two different combinations lead to the same result - as we had two combinations giving 40, the total number of answers is one less than this bound at 34.

Answer 3

I believe the answer is about 35 = 7c3 = the number of ways to choose 4 of four things when repeats are allowed, but let's do a quick count to make sure, because I think this may overcount at least one possibility.

Let's try grouping them by the number of pennies:
4 pennies: 4 (1 score)
3 pennies: 8, 13, 28 (3 scores)
2 pennies: 12, 17, 22, 32, 37, 52 (6 scores)
1 penny: 16, 21, 26, 31, 36, 41, 46, 56, 61, 76 (10 scores)
0 pennies: 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 80, 85, 100
(14 scores)

So the answer seems to be 34

Answer 4

+ redone;

This problem has been bugging me all day.

Add up different arrangements:

1. all the same - 4C1 = 4

2. a 1 and 3 combo of two scores
4C2 = 6, times 2 for reversing roles = 12

3.double double com: 4C2 = 6, now choose
the other 2, times 2 = 12

4. one double and 3 distinct: 4C3= 4 times 3
for the choice of which one to double = 12

5. all distinct 4C4 = 1 ways

adding them all
4+12+12+12+1 = 41 distinct total scores.

Answer 5

I shoot 25 (a million/2 a field) at each objective, a 6" "black" @ 25 yards.. I count variety the misses, no longer the hits. by potential of misses I propose "out of the black". I music my share hits. objective = a hundred% in a 6" "black" section. I often run approximately eighty 5% to 87% on 6-8 objectives with my centerfires on a sort day. hardly I hit a hundred% on a single objective. i might desire to artwork on that extra. With rimfire i will hit ninety 5%, and a hundred% on a single objective are plenty extra hardship-loose, yet i've got have been given a miracle gun for my .22.

Answer 6

4!=24

Answer 7

you can have,100,40,20.4,41,70,80,76,85,35,31.20.16.and on...and on......