Difference between revisions of "2001 AIME I Problems/Problem 6"
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=== Solution 2 === | === Solution 2 === | ||
+ | Let <math>a, b, c,</math> and <math>d</math> be the results of rolling the four dice respectively. We have the range <math>1\leq a\leq b\leq c\leq d\leq 6</math>, and there are <math>6^4=1296</math> total outcomes from rolling the dice. To transfer the inequality into a strictly increasing inequality, we can transform it into <math>1\leq a<b+1<c+2<d+3\leq 9</math>. Now, lets suppose <math>a'=a, b'=b+1, c'=c+2,</math> and <math>d'=d+3</math>. Then, <math>1\leq a'<b'<c'<d' \leq 9</math>. Clearly, there are <math>\binom{9}{4}=126</math> values that satisfy this equation, so our answer is <math>\dfrac{\binom{9}{4}}{6^4} = {\dfrac{7}{72}}\implies \boxed{079}</math>. | ||
+ | |||
+ | ~Mathkiddie | ||
+ | |||
+ | === Solution 3 === | ||
If we take any [[combination]] of four numbers, there is only one way to order them in a non-decreasing order. It suffices now to find the number of combinations for four numbers from <math>\{1,2,3,4,5,6\}</math>. We can visualize this as taking the four dice and splitting them into 6 slots (each slot representing one of {1,2,3,4,5,6}), or dividing them amongst 5 separators. Thus, there are <math>{9\choose4} = 126</math> outcomes of four dice. The solution is therefore <math>\frac{126}{6^4} = \frac{7}{72}</math>, and <math>7 + 72 = \boxed{079}</math>. | If we take any [[combination]] of four numbers, there is only one way to order them in a non-decreasing order. It suffices now to find the number of combinations for four numbers from <math>\{1,2,3,4,5,6\}</math>. We can visualize this as taking the four dice and splitting them into 6 slots (each slot representing one of {1,2,3,4,5,6}), or dividing them amongst 5 separators. Thus, there are <math>{9\choose4} = 126</math> outcomes of four dice. The solution is therefore <math>\frac{126}{6^4} = \frac{7}{72}</math>, and <math>7 + 72 = \boxed{079}</math>. | ||
− | === Solution | + | === Solution 4 === |
Call the dice rolls <math>a, b, c, d</math>. The difference between the <math>a</math> and <math>d</math> distinguishes the number of possible rolls there are. | Call the dice rolls <math>a, b, c, d</math>. The difference between the <math>a</math> and <math>d</math> distinguishes the number of possible rolls there are. | ||
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*If <math>a - d = 2</math>, then there are <math>{4\choose2} = 6</math> ways to arrange <math>b, c</math>, and there are only <math>6 - 2 = 4</math> values for <math>a, d</math>. | *If <math>a - d = 2</math>, then there are <math>{4\choose2} = 6</math> ways to arrange <math>b, c</math>, and there are only <math>6 - 2 = 4</math> values for <math>a, d</math>. | ||
− | Continuing, we see that the sum is equal to <math>\sum_{i = 0}^{5}{i+2\choose2}(6 - i) = 6 + 15 + 24 + 30 + 30 + 21 = 126</math>. The requested probability is <math>\frac{126}{6^4} = \frac{7}{72}</math>. | + | Continuing, we see that the sum is equal to <math>\sum_{i = 0}^{5}{i+2\choose2}(6 - i) = 6 + 15 + 24 + 30 + 30 + 21 = 126</math>. The requested probability is <math>\frac{126}{6^4} = \frac{7}{72}</math> and our answer is <math>\boxed{79}</math>. |
− | === Solution | + | === Solution 5 === |
The dice rolls can be in the form | The dice rolls can be in the form | ||
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<div style="text-align:center;"><math>\frac{m}{n} = \frac{15 + 60 + 15 + 30 + 6}{6^4} = \frac{126}{6^4} = \frac{7}{72}</math></div> | <div style="text-align:center;"><math>\frac{m}{n} = \frac{15 + 60 + 15 + 30 + 6}{6^4} = \frac{126}{6^4} = \frac{7}{72}</math></div> | ||
− | === Solution | + | Therefore, our answer is <math>\boxed{079}</math> |
+ | |||
+ | === Solution 6 === | ||
Consider the number of possible dice roll combinations which work after <math>1</math> roll, after <math>2</math> rolls, and so on. There is 6 possible rolls for the first dice. If the number rolled is a 1, then there are 6 further values that are possible for the second dice; if the number rolled is a 2, then there are 5 further values that are possible for the second dice, and so on. | Consider the number of possible dice roll combinations which work after <math>1</math> roll, after <math>2</math> rolls, and so on. There is 6 possible rolls for the first dice. If the number rolled is a 1, then there are 6 further values that are possible for the second dice; if the number rolled is a 2, then there are 5 further values that are possible for the second dice, and so on. | ||
Suppose we generalize this as a function, say <math>f(l,r)</math> return the number of possible combinations after <math>l</math> rolls and <math>r</math> being the beginning value of the first roll. It becomes clear that from above, <math>f(1,r) = 1</math>; every value of <math>l</math> after that is equal to the sum of the number of combinations of <math>l - 1</math> rolls that have a starting value of at least <math>r</math>. If we slowly count through and add up all the possible combinations we get <math>\frac{7}{72}</math> possibilities. | Suppose we generalize this as a function, say <math>f(l,r)</math> return the number of possible combinations after <math>l</math> rolls and <math>r</math> being the beginning value of the first roll. It becomes clear that from above, <math>f(1,r) = 1</math>; every value of <math>l</math> after that is equal to the sum of the number of combinations of <math>l - 1</math> rolls that have a starting value of at least <math>r</math>. If we slowly count through and add up all the possible combinations we get <math>\frac{7}{72}</math> possibilities. | ||
− | === Solution | + | === Solution 7 === |
In a manner similar to the above solution, instead consider breaking it down into two sets of two dice rolls. The first [[subset]] must have a maximum value which is <math>\le</math> the minimum value of the second subset. | In a manner similar to the above solution, instead consider breaking it down into two sets of two dice rolls. The first [[subset]] must have a maximum value which is <math>\le</math> the minimum value of the second subset. | ||
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Thus, the number of combinations is <math>\sum_{i=1}^6 i \cdot \left(\frac{(7 - i)(8 - i)}{2}\right) = 21 + 30 + 30 + 24 + 15 + 6 = 126</math>, and the probability again is <math>\frac7{72}</math>, giving <math>m+n=\boxed{079}</math>. | Thus, the number of combinations is <math>\sum_{i=1}^6 i \cdot \left(\frac{(7 - i)(8 - i)}{2}\right) = 21 + 30 + 30 + 24 + 15 + 6 = 126</math>, and the probability again is <math>\frac7{72}</math>, giving <math>m+n=\boxed{079}</math>. | ||
− | === Solution | + | === Solution 8 Recursion Formula === |
If you're too tired to think about any of the above smart transformations of the problem, a recursion formula can be a robust way to the correct answer. | If you're too tired to think about any of the above smart transformations of the problem, a recursion formula can be a robust way to the correct answer. | ||
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~Ilikeapos | ~Ilikeapos | ||
− | == Solution 11: | + | ===Solution 11: Stars and Bars(Under 2 minutes)=== |
− | + | Let the rolls be <math>a,b,c</math> and <math>d\newline</math> let <math>z=a-1, e=b-a, f=c-b, g=d-c, h=6-d\newline</math> <math>z+e+f+g+h=5\newline</math> This equation has <math>C(5+5-1, 5-1)=126</math> integer solutions<math>\newline</math> <math>126/1296=7/72\newline</math> <math>7+72=\boxed{79}</math> | |
− | + | ~ryanbear | |
+ | |||
+ | ===Solution 12=== | ||
− | ~ | + | Let's say the four dice values are all different. These can only be arranged in one way to satisfy our conditions, so there are <math>\binom{6}{4}=15</math> ways. If there are three different values, there are <math>\binom{6}{3}</math> to choose the numbers and <math>\binom{3}{1}</math> to choose which number will have the repeat, so <math>20\cdot3=60</math> ways. If there are two different values, there are <math>\binom{6}{2}</math> ways to choose the numbers. If there are two of each, then there is one way. If there are three of one and one of another, then there are <math>\binom{2}{1}</math> ways. Therefore <math>15\cdot(1+2)=45</math>. Now if all the numbers are the same, there are <math>\binom{6}{1}=6</math> ways. Altogether we have <math>15+60+45+6=126</math> ways. <math>\frac{126}{6^4}=\frac{21}{216}=\frac{7}{72}</math>. <math>7+72=\boxed{79}</math>. |
+ | ~MC413551 | ||
== See also == | == See also == |
Latest revision as of 03:16, 24 August 2024
Problem
A fair die is rolled four times. The probability that each of the final three rolls is at least as large as the roll preceding it may be expressed in the form where
and
are relatively prime positive integers. Find
.
Contents
Solutions
Solution 1
Recast the problem entirely as a block-walking problem. Call the respective dice . In the diagram below, the lowest
-coordinate at each of
,
,
, and
corresponds to the value of the roll.
The red path corresponds to the sequence of rolls . This establishes a bijection between valid dice roll sequences and block walking paths.
The solution to this problem is therefore . So the answer is
.
Solution 2
Let and
be the results of rolling the four dice respectively. We have the range
, and there are
total outcomes from rolling the dice. To transfer the inequality into a strictly increasing inequality, we can transform it into
. Now, lets suppose
and
. Then,
. Clearly, there are
values that satisfy this equation, so our answer is
.
~Mathkiddie
Solution 3
If we take any combination of four numbers, there is only one way to order them in a non-decreasing order. It suffices now to find the number of combinations for four numbers from . We can visualize this as taking the four dice and splitting them into 6 slots (each slot representing one of {1,2,3,4,5,6}), or dividing them amongst 5 separators. Thus, there are
outcomes of four dice. The solution is therefore
, and
.
Solution 4
Call the dice rolls . The difference between the
and
distinguishes the number of possible rolls there are.
- If
, then the values of
are set, and so there are
values for
.
- If
, then there are
ways to arrange for values of
, but only
values for
.
- If
, then there are
ways to arrange
, and there are only
values for
.
Continuing, we see that the sum is equal to . The requested probability is
and our answer is
.
Solution 5
The dice rolls can be in the form
ABCD
AABC
AABB
AAAB
AAAA
where A, B, C, D are some possible value of the dice rolls. (These forms are not keeping track of whether or not the dice are in ascending order, just the possible outcomes.)
- Now, for the first case, there are
ways for this. We do not have to consider the order because the combination counts only one of the permutations; we can say that it counts the correct (ascending order) permutation.
- Second case:
ways to pick 3 numbers,
ways to pick 1 of those 3 to duplicate. A total of 60 for this case.
- Third case:
ways to pick 2 numbers. We will duplicate both, so nothing else in this case matters.
- Fourth case:
ways to pick 2 numbers. We pick one to duplicate with
, so there are a total of 30 in this case.
- Fifth case:
; all get duplicated so nothing else matters.
There are a total of possible dice rolls.
Thus,
![$\frac{m}{n} = \frac{15 + 60 + 15 + 30 + 6}{6^4} = \frac{126}{6^4} = \frac{7}{72}$](http://latex.artofproblemsolving.com/f/a/7/fa7e76610f2e81a223ed2b61a6bcc7665e707692.png)
Therefore, our answer is
Solution 6
Consider the number of possible dice roll combinations which work after roll, after
rolls, and so on. There is 6 possible rolls for the first dice. If the number rolled is a 1, then there are 6 further values that are possible for the second dice; if the number rolled is a 2, then there are 5 further values that are possible for the second dice, and so on.
Suppose we generalize this as a function, say return the number of possible combinations after
rolls and
being the beginning value of the first roll. It becomes clear that from above,
; every value of
after that is equal to the sum of the number of combinations of
rolls that have a starting value of at least
. If we slowly count through and add up all the possible combinations we get
possibilities.
Solution 7
In a manner similar to the above solution, instead consider breaking it down into two sets of two dice rolls. The first subset must have a maximum value which is the minimum value of the second subset.
- If the first subset ends in a 1, there is
such subset and there are
ways of making the second subset.
- If the first subset ends in a 2, there is
such subsets and there are
ways of making the second subset.
Thus, the number of combinations is , and the probability again is
, giving
.
Solution 8 Recursion Formula
If you're too tired to think about any of the above smart transformations of the problem, a recursion formula can be a robust way to the correct answer.
We just need to work out the valid cases for each roll. Denote by the number of valid cases in the
-th roll when the number
is rolled, for
and
. Then we have the following recursion formula:
The logic is that, if
is rolled, then the number of valid cases is the subtotal of valid cases in the preceding roll for the outcome of 1 to
. The recursion can be easily calculated by hand when
are put in columns side by side, given the fact that te numbers are smaller than 100.
Finally, the total number of cases is
and
Solution 9: Observation of each case
Lets try casework and observe the cases. Notice that if the last roll is a , then the only dice rolls may be
, which is only
possibility. Observe that if the last roll is
, then there are
possibilities. When the last roll is a
, there are
possibilities. Notice when the last roll is
, the number of cases is the sum of the first
positive triangular numbers. This is easily provable when observing the numbers of possibilities after assigning a value to the last and second to last rolls.
So there are a total of
possibilities. So the probability is
and
. ~skyscraper
Solution 10: Distributions
This is equivalent to picking a four-element sequence of with repetition. Notice that once the sequence is picked, there is one and only one way to order these so that they form a sequence of rolls satisfying our conditions.
Now count the number of such four-element sequences, let be the number of
s in the sequence,
be the number of
s,
s,
s,
s, and
s. Now we see that we must have
with
being nonnegative integers since there are a total of
numbers picked. The number of solutions to this is
so our total number is equal to
making our answer
~Ilikeapos
Solution 11: Stars and Bars(Under 2 minutes)
Let the rolls be and
let
This equation has
integer solutions
~ryanbear
Solution 12
Let's say the four dice values are all different. These can only be arranged in one way to satisfy our conditions, so there are ways. If there are three different values, there are
to choose the numbers and
to choose which number will have the repeat, so
ways. If there are two different values, there are
ways to choose the numbers. If there are two of each, then there is one way. If there are three of one and one of another, then there are
ways. Therefore
. Now if all the numbers are the same, there are
ways. Altogether we have
ways.
.
.
~MC413551
See also
2001 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 5 |
Followed by Problem 7 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.