Dice Outcomes Setup
We are considering the scenario where three perfect dice, labeled \(D_1, D_2,\) and \(D_3\), are rolled simultaneously. Let the numbers appearing on these dice be represented by the variables \(x, y,\) and \(z\) respectively. A perfect die means each of its faces, numbered 1 through 6, has an equal chance of landing face up.
Condition \(x < y < z\) Analysis
The core requirement of the question is to find the number of possible outcomes \((x, y, z)\) where the numbers shown on the dice are strictly increasing. This means the number on the first die (\(x\)) must be less than the number on the second die (\(y\)), which must in turn be less than the number on the third die (\(z\)). Mathematically, this condition is expressed as:
\( x < y < z \)Since each die can show a number from 1 to 6, the possible values for \(x, y,\) and \(z\) are integers within the set \(\{1, 2, 3, 4, 5, 6\}\).
Key points about the condition \(x < y < z\):
- All three numbers (\(x, y, z\)) must be distinct. If any two were the same, the strict inequality (\(<\)) could not hold.
- The order matters in terms of which die shows which number (\(x\) on \(D_1\), \(y\) on \(D_2\), \(z\) on \(D_3\)), but the constraint \(x < y < z\) fixes the order once the numbers are chosen.
Combinations Calculation for Dice Rolls
The problem asks for the number of specific outcomes \((x, y, z)\) that satisfy \(x < y < z\). Because the condition \(x < y < z\) dictates a unique order for any set of three distinct numbers chosen, the problem simplifies to finding how many ways we can choose 3 distinct numbers from the set of possible outcomes on a single die, which is \(\{1, 2, 3, 4, 5, 6\}\).
This is a classic combinatorics problem. We need to select a subset of 3 numbers from a set of 6 numbers. The order in which we choose the numbers does not matter because once chosen, they can only be arranged in one way to satisfy \(x < y < z\). Therefore, we use the combination formula.
The formula for combinations, denoted as \(\binom{n}{k}\) or \(C(n, k)\), calculates the number of ways to choose \(k\) items from a set of \(n\) items without regard to the order of selection. The formula is:
\( \binom{n}{k} = \frac{n!}{k!(n-k)!} \)In this specific problem:
- \(n\) is the total number of possible outcomes for a single die, so \(n = 6\) (the numbers 1, 2, 3, 4, 5, 6).
- \(k\) is the number of distinct values we need to choose for \(x, y,\) and \(z\), so \(k = 3\).
Now, we apply the combination formula:
\( \binom{6}{3} = \frac{6!}{3!(6-3)!} \)First, calculate the factorials:
- \(6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720\)
- \(3! = 3 \times 2 \times 1 = 6\)
Substitute these values back into the formula:
\( \binom{6}{3} = \frac{720}{6 \times (6-3)!} = \frac{720}{6 \times 3!} = \frac{720}{6 \times 6} = \frac{720}{36} \)Alternatively, we can simplify the calculation:
\( \binom{6}{3} = \frac{6 \times 5 \times 4 \times 3!}{3! \times 3!} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} \) \( \binom{6}{3} = \frac{120}{6} \) \( \binom{6}{3} = 20 \)This result indicates that there are 20 distinct sets of three numbers that can be chosen from {1, 2, 3, 4, 5, 6}. Each of these sets corresponds to exactly one outcome where \(x < y < z\). For example, if the set {2, 4, 5} is chosen, the only outcome satisfying the condition is \((x=2, y=4, z=5)\).
Summary of Outcomes
The calculation shows that there are 20 possible combinations of numbers \((x, y, z)\) from rolling three dice such that \(x < y < z\). These combinations represent unique sets of three distinct numbers chosen from 1 to 6, arranged in ascending order.
Similar Questions
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Important Questions from Permutation and Combination
On a chess board, in how many different ways can 6 consecutive squares be chosen on the diagonals along a straight path ?
There are 6 persons arranged in a row. Another person has to shake hands with 3 of them so that he should not shake hands with two consecutive persons. In how many distinct possible combinations can the handshakes take place ?
In a tournament of Chess having 150 entrants, a player is eliminated whenever he loses a match. It is given that no match results in a tie/draw. How many matches are played in the entire tournament?
The letters A, B, C, D and E are arranged in such a way that there are exactly two letters between A and E. How many such arrangements are possible?
There is a numeric lock which has a 3-digit PIN. The PIN contains digits 1 to 7. There is no repetition of digits. The digits in the PIN from left to right are in decreasing order. Any two digits in the PIN differ by at least 2. How many maximum attempts does one need to find out the PIN with certainty?
