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# Let $x \cdot y \cdot z = 105$ where $x,y,z \in N$. Then number of ordered triplets $(x,y,z)$ satisfying the given equation isA) $15$B) $27$C) $6$D) $33$

Last updated date: 19th Jul 2024
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Hint: Here we are given to find the number of possible ordered triplets that can be formed such that product of those three values is equal to $105$. To do this we will use the method of prime numbers, where we write prime numbers with their powers, needed for the factorization of $105$. After finding the factors we find the possible triplets. Then we use those prime numbers to find other factors also and find the total possible triplets.
Formula used: Total numbers of possible ways to organize three things $a,b,b$ where two things are common is given by $\dfrac{{3!}}{{2!}}$.
Total numbers of possible ways to organize three things $a,b,c$ are $3!$.

Complete step-by-step solution:
Here we have to factorize $105$ using prime numbers. We can see that we can write $105$ as,
$105 = {3^1}{5^1}{7^1}$
Hence we get the factors as $3,5,7$. Now we have to find the possible triplets that can be formed by these numbers. We see that,
$3$ can be written at any place be it $x,y\,or\,z$. Same way numbers $5and7$ can also be written.
Hence, total possible triplets $= 3!$.
Same,
For $105 = 35 \times 3 \times 1$, we get possible triplets as $3! = 6$
For $105 = 21 \times 5 \times 1$, we get possible triplets as $3! = 6$
For $105 = 7 \times 15 \times 1$, we get possible triplets as $3! = 6$
For $105 = 105 \times 1 \times 1$, as two factors are same, we get possible triplets as $\dfrac{{3!}}{{2!}} = 3$
Hence we get possible triplets as, $6 + 6 + 6 + 6 + 3 = 27$