The total number of integral solutions of the | vedclass

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

Question

(B) First,find the number of positive integral solutions for $xyz = 24$.We write $24$ as $2^3 	imes 3^1$.The number of positive integral solutions is

Vedclass · Accepted Answer

$120$

Vedclass · Answer

(B) First,find the number of positive integral solutions for $xyz = 24$.We write $24$ as $2^3 	imes 3^1$.The number of positive integral solutions is given by the formula for distributing $n$ identical items into $r$ distinct bins,which is $\binom{n+r-1}{r-1}$.For $x = 2^{a_1} 3^{b_1}$,$y = 2^{a_2} 3^{b_2}$,$z = 2^{a_3} 3^{b_3}$,we have $a_1+a_2+a_3 = 3$ and $b_1+b_2+b_3 = 1$.The number of solutions for $a_i$ is $\binom{3+3-1}{3-1} = \binom{5}{2} = 10$.The number of solutions for $b_i$ is $\binom{1+3-1}{3-1} = \binom{3}{2} = 3$.Total positive integral solutions $= 10 	imes 3 = 30$.Since the product $xyz = 24$ is positive,the possible sign combinations for $(x, y, z)$ are $(+, +, +)$,$(+, -, -)$,$(-, +, -)$,and $(-, -, +)$.Each case has $30$ solutions.Total integral solutions $= 4 	imes 30 = 120$.

The total number of integral solutions of the equation $xyz = 24$ is

Similar Questions

If $^n{P_r} = 840$ and $^n{C_r} = 35$,then $n$ is equal to

Let $n = 1! + 4! + 7! + \ldots + 400!$. Then the ten's digit of $n$ is

If $\frac{1}{8!} + \frac{1}{9!} = \frac{x}{10!},$ find $x.$

The sum of all three-digit numbers,formed using non-zero digits,where all the digits are perfect squares of a natural number,is

The tens digit of the sum $1! + 4! + 7! + 10! + 12! + 13! + 16! + 17!$ is divisible by which of the following?

Vedclass Test Series

Exam Paper Generator

Online Exam Module