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Fastest way to compute this prob (BTG Expert-pls help)

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by voodoo_child » Sat Oct 08, 2011 7:00 am
Q - Three dices are rolled. For how many combinations will the sum of the numbers on the three dices is not greater than 13?

a) 162
b) 36
c) 160
d) 234
e) 158

Answer - 160

What's the best way? Any thoughts are greatly appreciated.
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by sl750 » Sat Oct 08, 2011 7:57 am
I don't think there is any simple way to solve this. You can reduce your burden by computing those combination of sums whose result is greater than or equal to 13 and subtracting this result from the total

There are 2 ways to obtain a sum greater than or equal to 13 in which all three digits are same
666
555
There are 30 ways to obtain a sum greater than or equal to 13 in which two digits are same
166
266
366
466
566
355
455
544
There are 24 ways to obtain a sum greater than or equal to 13 in which no two digits are same
256
346
356
456

Total ways = 216
Total ways of obtaining a sum greater or equal to 13 = 56 ways
Number of combinations in which the sum is no greater than 13 = 216-56 = 160
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by GmatMathPro » Sat Oct 08, 2011 9:03 am
I think this question is not good. The number of ways to roll three dice so that the sum is not greater than 13 should be 181. To get 160 you it would have to be "the sum is not greater than or equal to". Regardless, here is another method.

First the possible sums of two dice ranges from 2-12. Finding the pattern of distribution is fairly simple. There is one way to get 2, two ways to get 3, three ways to get 4.....six ways to get 7, five ways to get 8......one way to get 12.

Then, figure out how many options you have for the third die with each sum to satisfy the condition. Multiply these numbers together, and add them up. Here it is summarized in an excel spreadsheet. I'll do it with the restrictions the question writer apparently intended so you see that it matches up to the OA:

Image

Edit: hmmm...the chart's not displaying right. go to https://gmatmathpro.com/wp-content/uploads/dicechart.jpg to see the full chart.

Now just add the numbers in the last row: 6+10+12+12+10+6=56.

216-56=160

It's pretty easy once you see the pattern to the distributions. Might still be a faster way though.
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by voodoo_child » Sun Oct 09, 2011 4:09 pm
Pete - You are correct. That was a typo on my part :( I am sorry
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by GmatMathPro » Sun Oct 09, 2011 4:45 pm
Hey, no problem. Hope it helped!
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by Bek » Tue Oct 11, 2011 3:13 pm
There is one way to get 2, two ways to get 3, three ways to get 4.....six ways to get 7, five ways to get 8......one way to get 12.
Hi Pete, I am confused. Is there outcome of 2. If three dices are rolled, the minimum outcome is 3 not 2.

Please, explain.
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by GmatMathPro » Tue Oct 11, 2011 3:21 pm
GmatMathPro wrote: First the possible sums of two dice ranges from 2-12. Finding the pattern of distribution is fairly simple. There is one way to get 2, two ways to get 3, three ways to get 4.....six ways to get 7, five ways to get 8......one way to get 12.
Yes, I was talking about two dice. Two sentences before the one you quote I say "First the possible sums of two dice ranges from 2-12". My strategy was to first examine what can happen when you roll two dice and then see what happens when you add the third one in.
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by sohrabkalra » Wed Oct 12, 2011 4:12 am
The easiest i could think of is Total ways - Ways that give you More than 13( if you are talking not greater than 13)

To count the no of ways you can think of how to distribute the loss from 18 to the 3 digits

So ways to get
18 = 6-6-6 =1 [NO LOSS]
17 = 6-6-5 =1*3(ways of arrangement) =3 [ LOSS OF 1 ,distributed in 1-0-0]
16 = 6-6-4 =1*3 = 3 [LOSS OF 2 , distributed in 2-0-0]
= 6-5-5= 1*3 = 3 [LOSS OF 2, Distributed in 1-1-0]
15 = 6-6-3= 1*3 = 3 [Loss of 3, distributed in 3-0-0]
= 6-5-4= 1*6 = 6 [Loss of 3, distributed in 2-1-0]
= 5-5-5= 1*1 = 1 [Loss of 3, distribuuted in 1-1-1]
14 = 6-6-2= 1*3 = 3
= 6-5-3= 1*6 = 6
= 6-4-4= 1*3 = 3
= 5-5-4= 1*3 = 3

So No of ways of rolling dice so as not to get more than 13 = 216 - 35 =181

if you want to include 13,count for that too
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