One last difficulty: Peter rolls a fair 12-sided dice three times....

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yobacul

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Peter rolls a fair 12-sided dice three times. Each face of the dice has one of the numbers from 1 to 12 such that no face has the same number. The number on the top of the dice after each roll is recorded.

1) If the numbers are written down in the exact order in which they appear, how many different ordered triples are possible?

2) How many of the ordered triples would be made up of:
(a) the numbers 1, 1 and 2
(b) the numbers 1, 2 and 3

3) Hence, or otherwise, if the order in which the numbers are drawn is not important, how many different unordered triples are possible?

4) If the values of each triple are added and the sum obtained is recorded, how many different sums are possible?
 
Does the word 'ordered triple' mean consecutive numbers? Should I be checking for arrangements like (1, 2, 3) and what about repetitive digits? Is (1, 1, 2) also an ordered triple?
 
nad081 said:
Does the word 'ordered triple' mean consecutive numbers? Should I be checking for arrangements like (1, 2, 3) and what about repetitive digits? Is (1, 1, 2) also an ordered triple?
Click to expand...
That's a good thing to find out before trying to solve this problem! Have you tried looking up the term "ordered triple" to find a definition?

An ordered triple is a list of any three numbers, in which order matters (so it is not just a set of three numbers, and repetition is allowed as it would not be in a set).

You may be more familiar with "ordered pairs", such as the (x,y) used to identify a point on a coordinate plane. Here you are given two numbers, one representing the x-coordinate and the other the y-coordinate. They are a pair of numbers, in which order matters. In the same way, an ordered triple (x,y,z) locates a point in space.

In your context, an ordered triple just represents any possible result of rolling a die three times, namely the first, second, and third rolls, in that order.
 
So, for the first one I think it would be 12x12x12 different arrangements.

For the second one it would be a) 3!/2! =3 b) 3! =6
 
Yes, the number of possible ordered tripletons is [imath]12^3[/imath].

You can verify your reasoning in the second one as follows

With 1, 1, and 2, the 2 can be placed in any one of three positions, first, second, or third.

With 1, 2, and 3, the 1 can be placed in any one of three positions, and for each, there are two possible positions for the 2.
 
nad081 said:
Peter rolls a fair 12-sided dice three times. Each face of the dice has one of the numbers from 1 to 12 such that no face has the same number. The number on the top of the dice after each roll is recorded.
1) If the numbers are written down in the exact order in which they appear, how many different ordered triples are possible?
2) How many of the ordered triples would be made up of:
(a) the numbers 1, 1 and 2 (b) the numbers 1, 2 and 3
3) Hence, or otherwise, if the order in which the numbers are drawn is not important, how many different unordered triples are possible?
4) If the values of each triple are added and the sum obtained is recorded, how many different sums are possible?
Click to expand...
In addition to the comments in reply#4 you need help with your English Usage.
There are many forms of dice(plural). The cube is the most used form, but the regular octagonal shape die(singular) is a close second.
It has twelve faces numbered [imath]1\text[ to ]12[/imath] where opposite faces add to [imath]13[/imath]. There is no difference in tossing a die three times or tossing three of the same configured dice. We get the same set of outcomes.
Using a generating polynomial SEE HERE The term [imath]108x^{19}[/imath] tells us that there are [imath]108[/imath] ways to toss a sum equal to [imath]19[/imath].
Looking at the the exponents of terms, we see that possible sums range from [imath]3\text{ to }36.[/imath].
If we add the coefficients that sum should be [imath]1728\text{ or }12^3[/imath].
[imath][/imath]
 
Note: User "nad081" changed username to "yobacul".

Change approved.

Eliz.
 
JeffM said:
Yes, the number of possible ordered tripletons is [imath]12^3[/imath].

You can verify your reasoning in the second one as follows

With 1, 1, and 2, the 2 can be placed in any one of three positions, first, second, or third.

With 1, 2, and 3, the 1 can be placed in any one of three positions, and for each, there are two possible positions for the 2.
Click to expand...
So I managed the first two. Now, for 3) If the triplets are unordered would that just be 12C3?
 
You did not explain your reasoning. Probability problems require careful reasoning. You were given a subtle hint of one valid way to reason in this problem by the use of the word “hence.”

How many distinct unordered triples consist of all 1s?

So how many distinct unordered triples consist of the same number of pips?

How many distinct unordered triples consist of exactly two 1s and and one 2?

So how many distinct unordered triples consist of two different numbers of pips?

How many distinct unordered triples consist of exactly one 1, one 2, and one 3?

So how many distinct unordered triples consist of three different number of pips?

Putting that all together, how many distinct unordered triples are possible?

Does that equal 12C3?
 
Here is the definition of triple from Wolffram's MathWorld.
Frankly I have never seen a triple be any thing other than a set of three.
There are ordered triples, i.e. elements of [imath]S\times T\times U [/imath] for three sets.
The cardinality of which is: [imath]\#(S\times T\times U)=\#(S)\cdot\#(T)\cdot\#(U) [/imath]
 
c) You can either have the same number on each roll - 12C1 - 12
Else you can have two same numbers and one different - 12C2 x 11C1 -132
Or all the numbers will be different - 12C3 - 220

Hence total number of unordered triplets = 12+132+220=364

d) I figured it out
 
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