krishnasty wrote:
How many different combinations of outcomes can you make by rolling three standard (6-sided) dice if the order of the dice does not matter?
(a) - 24
(b) - 30
(c) - 56
(d) - 120
(e) - 216
First, if the dice were all unique, we could have calculated the number of combinations using basic counting principles i.e. we could have said that dice 1 can show 1/2/3/4/5/6 so 6 different outcomes, dice 2 has 6 different outcomes and dice 3 has 6 different outcomes. So possible outcomes = 6*6*6 = 216.
But we cannot do it here because the dice are identical. Also, I cannot say that since they are identical, let me just divide 216 by 3!. The reason is that not every possible outcome is counted 3! times when the dice are unique e.g.
_1_ _2_ _4_
D1 D2 D3
_1_ _4_ _2_
D1 D2 D3
and so on... Using 1, 2 and 4, we have 6 possible combinations.
But using 1, 2 and 2 we have only 3 possible combinations when the dice are unique and using 2, 2 and 2 there is only one possible combination. When we divide the entire 216 by 6!, we are reducing each of these combinations by a factor of 6 which is incorrect and that is why you get 216/6 = 36 i.e. less combinations than there actually are.
Then what to do in such questions?
Method 1: The better method is the one suggested above.
The solution of the dice can take three different forms.
Form 1: Each number is distinct e.g. 1, 2 and 4 above.
Out of 6 numbers (1/2/3/4/5/6), you select 3 using 6C3 = 20. You do not have to arrange these numbers since the dice are identical. So you will now get all such cases: (1,2, 3), (1, 2, 4), (1, 2, 5) ....(3, 4, 5)....(4, 5, 6)
Form 2: Two numbers are identical, one is different.
Out of 6 numbers (1/2/3/4/5/6), you select 1 which is going to be the identical one and then select another from the rest 5 numbers which will be the different one. So you get 6C1 * 5C1 = 30 combinations. Now you get all such cases: (1, 1, 2), (1, 1, 3), ....(2, 2, 1) ... (6, 6, 5)
Form 3: All three numbers are identical. Since there are 6 numbers, you can choose any one number in 6 ways. Now you will get all such cases: (1, 1, 1), (2, 2, 2)....(6, 6, 6)
Total possible combinations: 20 + 30 + 6 = 56
Worst case scenario: Let's say, it doesn't strike me at the time of the exam. The numbers are not so unwieldy that you cannot enumerate.
Just keep 'pattern recognition' in mind.
I say, I need to count all possible combinations.
(1, 1, 1), (1, 1, 2), (1, 1, 3) ... = 6
(1, 2, 2), (1, 2, 3)... = 5 (Here I do not have to count (1, 2, 1) because it is already taken above)
(1, 3, 3), (1, 3, 4) ... = 4
I see the pattern here. Numbers of the form (1, x, y) will be 6 + 5 + 4 + 3 + 2 + 1 = 21
Now (2, 2, 2), (2, 2, 3).... = 5 (I begin with 2, 2, 2 because I do not have to take a number lower than the one on the left i.e. I cannot take (2, 2, 1) since it has already been counted above as (1, 2, 2))
(2, 3, 3), (2, 3, 4) .... = 4
Again I see the pattern. Numbers of the form (2, x, y) will be 5 + 4 + 3 + 2 + 1 = 15
Which means numbers of the form (3, x, y) will be 4 + 3 + 2 + 1 = 10
Of the form (4, x, y) will be = 3 + 2 + 1 = 6
Of the form (5, x , y) will be = 2 + 1 = 3
Of the form (6, x, y) will be = 1
Here x is greater than or equal to the leftmost number and y is greater than or equal to x.
A painful method but do-able in 2 minutes if you recognize the pattern. To be used in worst case scenario.
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Karishma
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