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roulette

by mariah » Sun Feb 01, 2009 5:03 pm
82. The surface of a Clock (in a dart game) is evenly divided to eight parts numbered with numbers 1 to 8. If we throw arrows three times, how many ways are possible that the total score is 16?
pls explain
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Re: roulette

by piyush_nitt » Sun Feb 01, 2009 7:08 pm
mariah wrote:82. The surface of a Clock (in a dart game) is evenly divided to eight parts numbered with numbers 1 to 8. If we throw arrows three times, how many ways are possible that the total score is 16?
pls explain
IMO 45

not sure if there is any other way to solve this.

Basically we need to find out the number of ways in which we can make a sum = 16 using digits from 1-8.

Lets assume 1st digit is 8 then , no of ways

844
853
862
871
835
826
817

7 ways .

Now no of ways if hundreds digit : 7

772
763
754
745
736
727

6 ways

No of ways if hundreds digit : 6

664
655

2 ways

total number of watys = 15

there should be same number of arrangements when 8 ,7,6 is in middle or at unit's place

therefore 15*3 = 45

OA?
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by Alara533 » Sun Feb 01, 2009 8:47 pm
Total number is 42.

Piyush counted 6 combination for 772, where as it has only 3 combination.

Btw, shouldn't we count for the case where the dart missed the target, ie, 0 points...8,8,0? Its not mentioned that the board was hit all the 3 times!

In that case, we get an extra 3 combination to make it 45 :).
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by ontopofit » Sun Feb 01, 2009 11:42 pm
IMO 54

we shud form diff sets of number summing up to 16
we start from 8......also the arrow can hit the same number twice
871
862
853
844.......so we have 4*3! ways with this

taking 7
now we shud make sure that we dont have to include 8 this time
772
763
754.......3*3! ways

taking 6 (not 7 or 8)
664
655.........2*3! ways


so total number of ways = 24 + 18 + 12 = 54


can u post OA plz?
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by drpawan » Mon Feb 02, 2009 1:36 am
I agree with ontopofit....
My ans would be 54..
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by krisraam » Mon Feb 02, 2009 8:21 am
ontopofit wrote:IMO 54

we shud form diff sets of number summing up to 16
we start from 8......also the arrow can hit the same number twice
871
862
853
844.......so we have 4*3! ways with this

taking 7
now we shud make sure that we dont have to include 8 this time
772
763
754.......3*3! ways

taking 6 (not 7 or 8)
664
655.........2*3! ways


so total number of ways = 24 + 18 + 12 = 54


can u post OA plz?
The ways 844,772,655 are 3(3!/2!) each not 3!.

So the answer is 45.

Thanks
Raama
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by Alara533 » Mon Feb 02, 2009 9:19 am
Raama, same with 664 right? Its 3!/2! and not 3!
So won't it further reduce to 42?
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by krisraam » Mon Feb 02, 2009 10:28 am
Alara533 wrote:Raama, same with 664 right? Its 3!/2! and not 3!
So won't it further reduce to 42?
Yes Alara533. Thanks for pointing it out. I missed it.

Thanks
Raama
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by mariah » Sat Feb 07, 2009 3:31 pm
oa 54
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by mariah » Sun Feb 08, 2009 8:01 pm
i wrote down all comb. and its 42

guys, i cant follow your expl :roll:
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by Uri » Mon Feb 09, 2009 2:29 am
mariah wrote:i wrote down all comb. and its 42

guys, i cant follow your expl :roll:
Ontopofit has shown the correct procedure, with some small computational mistakes, which have been later rectified by krisraam and Alara533. I am copying the entire solution of Ontopofit and incorporating the necessary rectifications in it.


we shud form diff sets of number summing up to 16
we start from 8......also the arrow can hit the same number twice
871 -> 3! ways = 6 ways
862 -> 3! ways = 6 ways
853 -> 3! ways = 6 ways
844 -> 3!/2! ways = 3 ways
----------------------------------
sub-total= 21 ways


taking 7
now we shud make sure that we don't include 8 this time
772 -> 3!/2! ways = 3 ways
763 -> 3! ways = 6 ways
754 -> 3! ways = 6 ways
----------------------------------
sub-total= 15 ways

taking 6 (not 7 or 8)
664 -> 3!/2! ways = 3 ways
655 -> 3!/2! ways = 3 ways
----------------------------------
sub-total= 6 ways

so, total= 21+15+6= 42 ways

The funda of this method is that we count the ways we hit a specific number and then hit other relevant numbers so that the total becomes 16. Once we consider hitting 8, we find the other necessary numbers to make the total 16 (8,7,1; 8,6,2; 8,5,3; 8,4,4). Now, we can hit 8 in the first throw, in the second throw or in the third throw. So, we find the number of ways that we make 16 with 8 in it and any particular combination that we have found.
Let us consider 8,7,1. We can get this combination in 6 ways.
While considering 8,4,4, we see that we can get this combination in 3 ways. This difference is due to the repetition.
This same logic applies to all the combinations.
Now, when we consider hitting 7, we should not consider 8 as any element of the combination, as we have already counted that while considering 8 earlier. Same applies for 6. And we see that we don't have any unique combination left with 5 or any numbers below that, which we have not still considered.
So, we proceed to find the total which is 42.
Hope this helps.
Since your OA (whatever source it is) is 54, I would like to know whether they have provided any explanation. Can you also mention the source?
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by mariah » Mon Feb 09, 2009 6:04 am
i have collected it from net. dont know exact source
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