How many different combinations of outcomes can you make by

C 56

I count all the possible combo.

definately not 6*6*6 since the question said 'order does not matters'.
It means we do not distinguish between (1,1,5) or (1,5,1) or (5,1,1) .All these are considered as 1 outcome.

Some further explanation on my answer above:

Case (2) is different from case (3).
In case 2, e.g. 5, (6, 6) and 6, (5, 5) are different.

However, in case (3), e.g. 2, 3, 4, and 4, 2, 3 are the same.

So, in case 2, you don't divide it by 2!, but in case (3), you divide it by 3!.
Case (2) is more of a permutation on itself.

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.

The only way I can see this is 6*6*6 - 6 (same outcomes) ... Can anyone explain?

HI BUnuel,

Could you please explain me this part --

2. XXY - two dice show alike numbers and third is different: \(6*5=30\), 6 choices for X and 5 choices for Y;

i got 1st and 3rd but this one..

Regards
Ishdeep Singh

Would be easier to list all possible cases:
XXY
112,
113,
114,
115,
116,

221,
223,
224,
225,
226,

331,
332,
334,
335,
336,

441,
442,
443,
445,
446,

551,
552,
553,
554,
556,

661,
662,
663,
664,
665.

6 choices for X and 5 choices for Y.

Does this make sense?

Hi B,

2. XXY - two dice show alike numbers and third is different: \(6*5=30\), 6 choices for X and 5 choices for Y;
3. XYZ - all three dice show distinct numbers: \(C^3_6=20\), selecting three different numbers from 6;

Could you explain why do we take combination for 3rd choice 3)- \(C^3_6\) and not for the 2nd choice 2) \(C^2_6\)?

Thanks,
PK

For the third case (XYZ) \(C^3_6=20\) gives all 3-number combinations out of 6, which is exactly what we need for this case:
{1, 2, 3}
{1, 2, 4}
...
{4, 5, 6}

While if we use \(C^2_6=15\) for the second case (XXY) we get all 2-number combinations out of 6:
{1, 2}
{1, 3}
{1, 4}
...
{5, 6}

But we are rolling 3 dice and we need to take this into account. For example, {1, 2} case can be {1, 1, 2} or {1, 2, 2} and the same for all other cases, which means that we need to multiply \(C^2_6=15\) by 2 to get the total number of this case.

Does this make sense?

Hi Bunuel

Could you please show how to go about working on the same problem if the ORDER of the dice mattered.
What would the approach be?

Thanks in advance.

Order of the dice matters means the dice are distinct. Think of having 3 dice of 3 different colors - Red, Blue and Yellow. So (5, 1, 1) is not 1 case now but 3 cases since Red die could show 5 or Yellow could show 5 or Blue could show 5.

Now its like 3 distinct places each of which can take 6 distinct values (1/2/3/4/5/6). So total number of outcomes = 6*6*6 = 216

If some of you are like me and find it difficult to think in combinations, I started by thinking about the possible outcomes, in order.

111, 112, 113, 114, 115, 116 = 6 possibilities

121 has already been counted with 112, so we notice that every digit goes in increasing order. Thus the next row has to start with 122 instead of 121.

122, 123, 124, 125, 126 = 5 possibilities.

Continuing on, we notice that our total number can be given by:

Starts with 1: 6+5+4+3+2+1 possibilities (111 - 116, 122-126, 133-136,....,166)
Starts with 2: 5+4+3+2+1 possibilities (222-226, 233-236, 244-246, 255-256, 266)
Starts with 3: 4+3+2+1 possibilities (333-336, 344-346,355-356,366)
Starts with 4: 3+2+1 possibilities(444-446, 455-456,466)
Starts with 5: 2+1 possibilities (555-556, 566)
Starts with 6: 1 possibility = 666

Thus, our total will just be the sum of 1+3+6+10+15+21 = 56.

Answer: C

Dear Bunnel

SO basically, when the qs stem says different combination outcomes when order doesnt matter it means that we have to take all possibilities in account:

XXX: i understood: 111, 222, 333, 444, 555, 666
XXY: what about XYX & XXY??? we have not counted this as qs says order doesnt matter?
XYZ: we have just counted 6*5*4/3! as order doesnt matter??

How do you get to know what the qs is asking?
when i read the qs I thought it was asking for all possible outcomes when three dice are thrown. so my answer was 6*6*6 = 216.
what is the difference between order doesnt matter and 216?

Yes, since the order of the dice does not matter then XXY, XYX and YXX are the same for this problem. Similarly, XYZ, XZY, YXZ, ... are also the same.

and when would it be 6*6*6??

When the order matters, so when {1, 1, 2} is different from {1, 2, 1} and {2, 1, 1}.

Let me add my thought too here: When order matters, essentially you are saying that the dice are distinct, say of three different colors. So a 2 on the red die and 1 on the others is different from 2 on the yellow one with 1 on the other two.
When order doesn't matter, it implies that the dice are identical. If you have three identical dice and you throw them, a 2, 1, 1 is the same no matter which die gives you 2 because you cannot tell them apart.

Hi
Can I get any link which make me easy to understand this topic boz I m very much confused.
Kindly help me combination and selection are very confusing topics.
Thx

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