The number of triplets (x, y, z),where x, y, | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The total number of non-negative integer solutions to $x+y+z=15$ is given by the stars and bars formula: $\binom{15+3-1}{3-1} = \binom{17}{2} = \f

Vedclass · Accepted Answer

$114$

Vedclass · Answer

(B) The total number of non-negative integer solutions to $x+y+z=15$ is given by the stars and bars formula: $\binom{15+3-1}{3-1} = \binom{17}{2} = \frac{17 	imes 16}{2} = 136$.Let $S$ be the set of all non-negative integer solutions. We want to find the number of solutions where $x, y, z$ are distinct.First,find the number of solutions where at least two variables are equal:Case $1$: $x=y$. Then $2x+z=15$. Since $z \ge 0$,$2x \le 15$,so $x \in \{0, 1, 2, 3, 4, 5, 6, 7\}$. This gives $8$ solutions.By symmetry,the cases $y=z$ and $x=z$ also yield $8$ solutions each.Total solutions with at least two variables equal is $8+8+8 = 24$.Case $2$: $x=y=z$. Then $3x=15$,so $x=5$. This gives $1$ solution $(5, 5, 5)$.Note that the solution $(5, 5, 5)$ is counted in all three sub-cases of Case $1$ ($x=y$,$y=z$,$x=z$).Using the Principle of Inclusion-Exclusion,the number of solutions where at least two are equal is $8+8+8 - 2(1) = 22$.Number of solutions where $x, y, z$ are distinct = (Total solutions) - (Solutions where at least two are equal) = $136 - 22 = 114$.

The number of triplets $(x, y, z)$,where $x, y, z$ are distinct non-negative integers satisfying $x+y+z=15$,is

Similar Questions

The number of $4$-digit numbers that can be formed from the digits $0, 1, 2, 3, 4, 5, 6, 7$ such that each number contains the digit $1$ is:

The number of integers lying between $1000$ and $10000$ such that the sum of all the digits in each of those numbers becomes $30$ is:

Total number of $3$-digit numbers that are divisible by $6$ and can be formed by using the digits $1, 2, 3, 4, 5$ with repetition is $.......$.

For a natural number $n$,the inequality $2^n(n - 1)! < n^n$ holds true if:

The number of ways in which $4$ different things can be distributed to $6$ persons so that no person gets all the things is

Vedclass Test Series

Exam Paper Generator

Online Exam Module