There are n sets of numbers each having only three positive integers with LCM equal to 1001 and HCF equal to 1. What is the value of n?

The correct option is More than 8.

Let's find the value of $n$, which represents the total number of sets of three positive integers $(a,b,c)$ such that:

1. $\mathrm{LCM}(a,b,c)=1001$
2. $\mathrm{HCF}(a,b,c)=1$

First, let's find the prime factorization of 1001:
$1001=7\times 11\times 13$

Let the three numbers be of the form:
$a=7^{p_1}\times {11}^{q_1}\times {13}^{r_1}$
$b=7^{p_2}\times {11}^{q_2}\times {13}^{r_2}$
$c=7^{p_3}\times {11}^{q_3}\times {13}^{r_3}$

Where each exponent $p_i,q_i,r_i\in \{0,1\}$.

For each prime factor $p\in \{7,11,13\}$:

- Since $\mathrm{LCM}(a,b,c)=1001$, the maximum exponent among the three numbers for each prime factor must be 1.
- Since $\mathrm{HCF}(a,b,c)=1$, the minimum exponent among the three numbers for each prime factor must be 0.

Thus, for each prime factor, the exponents chosen for $(a,b,c)$ must be a combination of 0s and 1s that contains at least one 0 and at least one 1.
Out of all $2^3=8$ possible combinations of 3 binary exponents (0 or 1):
- The combination $(0,0,0)$ is invalid (maximum is not 1).
- The combination $(1,1,1)$ is invalid (minimum is not 0).

So, there are $8-2=6$ valid exponent choices for each prime factor.

Since the choices for each prime factor (7, 11, and 13) are independent, the total number of ordered triplets $(a,b,c)$ is:
$6\times 6\times 6=216$

To find the total number of unordered sets of 3 distinct positive integers, we must account for symmetric triplets:

1. Triplets with 3 distinct numbers:
Each distinct set $\{a,b,c\}$ can be ordered in $3!=6$ ways.

2. Triplets where two numbers are equal (e.g., $a=b\ne c$):
For $a=b$, the exponents for each prime factor must be identical for $a$ and $b$, and different for $c$.
The exponent pairs $(p_1,p_1,p_3)$ must have one value 0 and the other 1:
- $(0,0,1)$ or $(1,1,0)$ -> 2 valid choices per prime factor.
So there are $2^3=8$ triplets of the form $(a,a,c)$.
Similarly, there are 8 triplets for $(a,b,a)$ and 8 triplets for $(b,a,a)$.
Thus, there are $3\times 8=24$ ordered triplets having 2 equal numbers.

Subtracting these from the total gives the number of ordered triplets with 3 distinct elements:
$216-24=192$

The number of distinct unordered sets of 3 numbers is:
$\frac{192}{6}=32$

Since 32 is significantly greater than 8, the value of $n$ is More than 8.