So, as per the question, we're choosing six real numbers
namely, $X_1,X_2,X_3,X_4,X_5,X_6$
Now, I'll write $6$ equations that should be satisfied the criteria:
The product of any five of them is equal to other number.
$$X_1.X_2.X_3.X_4.X_5=X_6 ---- A) \\ X_1.X_2.X_3.X_4.X_6=X_5 ---- B) \\ X_1.X_2.X_3.X_5.X_6=X_4 ---- C) \\ X_1.X_2.X_4.X_5.X_6=X_3 ---- D) \\ X_1.X_3.X_4.X_5.X_6=X_2 ---- E) \\ X_2.X_3.X_4.X_5.X_6=X_1 ---- F)$$
Now, take A) & B) & divide them
$\dfrac{A)}{B)} $ $:$ $\dfrac{X_1.X_2.X_3.X_4.X_5}{X_1.X_2.X_3.X_4.X_6} = \dfrac{X_6}{X_5}$
Or, $\dfrac{A)}{B)}$ $:$ $\dfrac{X_5}{X_6} = \dfrac{X_6}{X_5}$
Or, $(X_5)^2 = (X_6)^2$
take A) & C) & divide them,
$\dfrac{A)}{C)} $ $:$ $\dfrac{X_1.X_2.X_3.X_4.X_5}{X_1.X_2.X_3.X_5.X_6} = \dfrac{X_6}{X_4}$
Or, $\dfrac{A)}{B)}$ $:$ $\dfrac{X_4}{X_6} = \dfrac{X_6}{X_4}$
Or, $(X_4)^2 = (X_6)^2$
∴ If we take any pair from A), B), C), D), E), F) and divide one by other, we'll get -
$$ (X_1)^2 = (X_2)^2 = (X_3)^2 = (X_4)^2 = (X_5)^2 = (X_6)^2 $$
∴ All the numbers have the same magnitude, only difference will be their sign.
Let's assume the magnitude of each of the real numbers is $\pm X$.
So, the arrangement of plus or minus signs is going to determine the arrangements possible there.
And we have to find how we'll get
$$X.X.X.X.X = X \\ X^5 = X$$
Only when $x =1$ or $X=0$, then $X^5 = X$
But, there could be an another way to get $X^5 = X$
When there is even number of $-1's$ (i.e. $2$ or $4$ only), the product of $5 x$ will be $1$ which is equal to the selected $x$ (i.e. $1$)
[number of $-1's$ can't be $6$ because, product of $5$ real no.s should be equal to the $6^{th} $ no.(other number)]
$$-1.-1.1.1.1 = 1 \\ \text{ Or} \\ -1.-1.-1.-1.1 =1 $$
Now, if we switch to the binary system where
$$\text{0 represents +} \\ \text{1 represents -}$$
Then we have $2^6 = 64$ combinations of plus & minus
but, we only want those combination where we have only even no. of $-1's$
We can apply the parity concept.
Now, we have only $32$ combinations of bits with even parity.
So, we have $32$ arrangements for the $x's$ with their signs which satisfy the criteria or the required conditions.
plus we have one more solution where all $x's$ are $0$.
∴ $\color{green}{\text{Total arrangement or choices will be} = 32+1}$ = $\color{gold}{33}$
& this can think of-
1) Where all $x's$ are $0$, product of $5$ $x's$ will be $0$, which is equal to another $x$ (i.e. $0$)
2) Where all $x's$ are $+1$, product of $5$ $x's$ will be $+1$, which is equal to another $x$ (i.e. $+1$)
3) Where all $x's$ are $-1$
$$\text{think as, } -1.-1.-1.-1.-1 = -1 $$
4) Where $4$ $x's$ are $+1$ & $2$ $x's$ will be $-1$
$$\text{think as, } +1.+1.+1.+1.-1 = -1$$
5) Where $4$ $x's$ are $-1$ & $2$ $x's$ will be $+1$
$$\text{think as, } -1.-1.-1.-1.+1 = +1$$
1) can be done in = $1$ ways $\left [ \dfrac{6!}{6!} = 1 \right]$
2) can be done in = $1$ ways $\left [ \dfrac{6!}{6!} = 1 \right]$
3) can be done in = $\dfrac{6!}{6!} = 1 ways$
4) can be done in = $\dfrac{6!}{4! \times 2!} = \dfrac{6 \times 5}{2} = 15\text{ways}$
5) can be done in $\dfrac{6!}{4! \times 2!} = \dfrac{6 \times 5}{2} = 15\text{ways}$
∴ $\color{green}{\text{Total choices }}$= $1) + 2) + 3) + 4) + 5)$
$\qquad \qquad = 1+1+1+15+15$
$\qquad \qquad = \color{gold}{33 \text{ choices }}$