Let $G$={$1, -1$} be the group with multiplication and $H= G \times G \times G$ be the group with the operation defined for $x_1=(a_1, b_1, c_1)$ , $x_2=(a_2, b_2, c_2)$ as $x_1 * x_2 = (a_1a_2, b_1b_2, c_1c_2)$.
How many subgroups are there in H with order $4$ ?
My answer is $7$ but I think there could be some wrong calculations.
Note that for any $h \in H$, we have $h^2= 1_H$, and $H$ is abelian. Therefore any subgroup of order $4$ has the form $\{1,a,b,ab\}$ for elements $a, b \in H- \{1\}$. There are $\binom 72$ possiblities to choose $2$ elements from $H$, but every subgroup is counted $\binom 32$-times in this way, because we get the same subgroup if we choose $2$ of the 3 non-units of $\{1,a,b,ab\}$. Therefore there are $$ \frac{\binom 72}{\binom 32} = \frac{21}{3} = 7 $$ subgroups of order 4, as you state correctly.