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Since the total score is 15 and the possible scores are 2, 5, and 10, one of the scores must be 5 (you can use the even odd concept, but here it is clearly evident).
=> The rest of the sum must be 10.
1) 10 can be obtained in only way using 10 directly. So, the scores becomes [5,10,0,0,0,0] = 6!/4! combinations = 30
2) 10 can be obtained using 5 2s. Possible combination [5,2,2,2,2,2] = 6 combinations
3) 10 can be obtained using 2 5s. Possible combination [5,5,5,0,0,0] = 6!/(3!*3!) = 20

Total: 20+30+6=56

Ans: B
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Points on hitting inner most circle = 10
Points on hitting outer circle excluding inner circle = 5
Points on hitting outer most circle = 2

in 6 shots = 15 points

n*10 + m*5 + q*2 = 15

1) In first case, its possible that ( 2,2,2,2,2,5) [ Total points = 15]
Total no. of ways = 6C1 = 6
2) In second case, its possible that ( 5,5,5,,0,0,0) [ Total points = 15]
Total no. of ways = 6C3 = 20
3) In third case, its possible that ( 10,5,0,0,0,0) [ Total points = 15]
Total no. of ways = 6C2 * 2! = 30

Total no. of ways = 30+20+6 = 56

Answer: B
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A dart board is made of three concentric circles such that hitting the area of the innermost circle will give 10 points, hitting the area of the second circle excluding the area of the innermost circle will give 5 points and hitting the area of the outermost circle excluding the area of the second circle will give 2 points. If in 6 shots 15 points have been earned then in how many ways could that have been done?

the points that can be earned are 5, 2 and 10

in 6 shots 15 points .

Case 1 :
10 , 5 ,0 ,0,0,0 .

No of ways = 6! / 4! = 30

Case 2 :

5,5,5,0,0,0

No of ways = 6! / (3! * 3!)= 20

Case 3 :

5 ,2,2,2,2,2

No of ways = 6! / (5!)= 6



Total no of ways 20+30+6=56

B is the ans .
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Kindly see the attachment.
B = 56
Attachments

1.PNG
1.PNG [ 47.03 KiB | Viewed 6474 times ]

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Bunuel, are you sure the prompt is complete? Shouldn't it mention that missing the board means 0 points?
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