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
- A man has \(7\) relatives (\(4\) women and \(3\) men). His wife also has \(7\) relatives (\(3\) women and \(4\) men). In how many ways can they invite \(3\) women and \(3\) men so that \(3\) of them are man's relatives and \(3\) of them are his wife's relatives?
- A triangle \(PQR\) is such that \(3\) points lie on the side \(PQ\), \(4\) points on \(QR\) and \(5\) points on \(RP\) respectively. Triangles are constructed using these points as vertices. What is the number of triangles so formed ?
- How many four-digit natural numbers are there such that all of the digits are even?
- How many 4-digit numbers are there having all digits as odd?
- In how many ways can the letters of the word INDIA be permutated such that in each combination, vowels should occupy odd positions?
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- What is the sum of all four digit numbers formed by using all digits \(0, 1, 4, 5\) without repetition of digits?
- In how many ways can a student choose \((n-2)\) courses out of \(n\) courses if 2 courses are compulsory \((n > 4)\)?
If the number of selections of r as well as (n + r) things from 5n different things are equal, then what is the value of r?
Important Questions from Permutation and Combination
The following figure is made with nine squares of length 1 cm each.
A student is standing at point P. The total number of shortest paths to reach point Q from point P, if the student walks along the boundaries of the squares only, is:
Consider the following statements (all the angles are measured in degrees) :
1. If two supplementary angles are $4x + 40$ and $10x$, then $x$ is;
2. The complementary angle of $\frac{4}{5} \times 45$ is :
3. If two complementary angles are in the ratio of $4 : 5$, then the smallest angle is :
Which of the following is correct?
Codes:
1 2 3- There are 40 persons in a palace. If every person shakes hands with every other person, what will be the total number of handshakes?
- How many words can be formed with or without meaning by using all letters of the word DELHI?
A rectangular wall is divided into four squares of equal size where there are two rows each having two squares. The top left square is coloured with green. If, including green, there are three colours available and each square is coloured using any one of these three colours such that no two adjacent squares get painted with the same colour; then how many colour combinations are possible ?
