7 Saints, 7 Halos, and 700 Images ( Not Original )

Let these boxes with numbers represent the seven saints and the number of halos above each one.

$\Large \boxed{ \bigcirc } \, \boxed{\bigcirc} \, \boxed{\bigcirc} \, \boxed{\bigcirc} \, \boxed{\bigcirc} \, \boxed{\bigcirc} \, \boxed{\bigcirc}$

Let this string of boxes and 1’s represent the number of halos over each saint.

$\Large 1 \boxed{ \bigcirc } 1 \boxed{ \bigcirc} 1 \boxed{\bigcirc} 1 \boxed{\bigcirc} 1 \boxed{\bigcirc} 1 \boxed{\bigcirc} 1$

For $7$ saints with halos $\left( x = 7 \right)$ , we must choose all $6$ boxes to place a $+$ sign to form the sum

$\Large 1 \boxed{+} 1 \boxed{+} 1 \boxed{+} 1 \boxed{+} 1 \boxed{+} 1 \boxed{+} 1 = 1+1+1+1+1+1+1$

Thus, there is only $1$ image that can be formed where all $7$ saints have a halo.

For $x = 6$ ( $1$ saint without a halo ), we must choose $5$ of $6$ boxes to place a $+$ sign to form the sums of $6$ saints sharing the $7$ halos. This can be done in ${6 \choose 5} = 6$ ways:

$\begin{matrix} &1 \boxed{\bigcirc} 1 \boxed{+} 1 \boxed{+} 1 \boxed{+} 1 \boxed{+} 1 \boxed{+} 1 &= 2+1+1+1+1+1 \\ &1 \boxed{+} 1 \boxed{\bigcirc} 1 \boxed{+} 1 \boxed{+} 1 \boxed{+} 1 \boxed{+} 1 &= 1+2+1+1+1+1 \\ &1 \boxed{+} 1 \boxed{+} 1 \boxed{\bigcirc} 1 \boxed{+} 1 \boxed{+} 1 \boxed{+} 1 &= 1+1+2+1+1+1 \\ &1 \boxed{+} 1 \boxed{+} 1 \boxed{+} 1 \boxed{\bigcirc} 1 \boxed{+} 1 \boxed{+} 1 &= 1+1+1+2+1+1 \\ &1 \boxed{+} 1 \boxed{+} 1 \boxed{+} 1 \boxed{+} 1 \boxed{\bigcirc} 1 \boxed{+} 1 &= 1+1+1+1+2+1 \\ &1 \boxed{+} 1 \boxed{+} 1 \boxed{+} 1 \boxed{+} 1 \boxed{+} 1 \boxed{\bigcirc} 1 &= 1+1+1+1+1+2 \end{matrix}$

In general if we want to have $x$ saints sharing $7$ halos we will have ${ 6 \choose x-1 }$ ways to do this:

${ 6 \choose x-1 } = \frac{6!}{ (x-1)! \,(7-x)! } \quad \text{Eq1}$

Now, ( to start with ) distribute the halos to the saints choosing the farthest saint on the left to NOT have a halo:

$\begin{matrix} &\boxed{\bigcirc} \, &\boxed{2} \, &\boxed{1} \, &\boxed{1} \, &\boxed{1} \, &\boxed{1} \, &\boxed{1} \\ &\boxed{\bigcirc} \, &\boxed{1} \, &\boxed{2} \, &\boxed{1} \, &\boxed{1} \, &\boxed{1} \, &\boxed{1} \\ &\boxed{\bigcirc} \, &\boxed{1} \, &\boxed{1} \, &\boxed{2} \, &\boxed{1} \, &\boxed{1} \, &\boxed{1} \\ &\boxed{\bigcirc} \, &\boxed{1} \, &\boxed{1} \, &\boxed{1} \, &\boxed{2} \, &\boxed{1} \, &\boxed{1} \\&\boxed{\bigcirc} \, &\boxed{1} \, &\boxed{1} \, &\boxed{1} \, &\boxed{1} \, &\boxed{2} \, &\boxed{1} \\&\boxed{\bigcirc} \, &\boxed{1} \, &\boxed{1} \, &\boxed{1} \, &\boxed{1} \, &\boxed{1} \, &\boxed{2} \end{matrix}$

Choosing the first saint to NOT have one of the halos was arbitrary. Any one of the $7$ saints could be chosen for a total of ${ 7 \choose 1 } = 7$ ways. In general, when we want $7-x$ saints to not have a halo we will have ${ 7 \choose 7-x }$ ways to do this :

${ 7 \choose 7-x } = \frac{7!}{ (7-x)! \, x! } \quad \text{Eq2}$

Thus, the number of images $N$ that can made for some number $x$ saints that have a halo is given by the product ( number of ways that $x$ saints can share $7$ halos times the number of ways to choose the $\left( 7-x \right)$ without halos out of $7$ saints ):

$N = { 6 \choose x-1 } \cdot { 7 \choose 7-x } = \frac{6!}{ (x-1)! \,(7-x)! } \cdot \frac{7!}{ (7-x)! \, x! }$

For this problem we are trying to find $x$ such that $N = 700$ :

$700 = \frac{6!}{ (x-1)! \,(7-x)! } \cdot \frac{7!}{ (7-x)! \, x! } = \frac{6! \, 7!}{ {(7-x)!}^2 \, (x-1)! \, x! } \quad \text{Eq3}$

I don’t know of any way to form analytical solutions to $\text{Eq3}$ , so I used guess and check:

For $x = 4$ we have:

$\frac{6! \, 7!}{ {(3)!}^2 \, (3)! \, 4! } = \frac{ \cancel{3!} \cdot 5 \cdot 4 \cdot \cancel{3!} \cdot 7 \cdot \cancel{3!} \cdot 5 \cdot \cancel{4!} }{ \cancel{ {3!}^3 } \cdot \cancel{4!} } = 7 \cdot 5^2 \cdot 4 = 700$