## Pascal’s Triangle Combinations – product of 6 integers

$\dbinom{0}{0}$

$\dbinom{1}{0}$   $\dbinom{1}{1}$

$\dbinom{2}{0}$   $\dbinom{2}{1}$   $\dbinom{2}{2}$

$\dbinom{3}{0}$   $\dbinom{3}{1}$   $\dbinom{3}{2}$   $\dbinom{3}{3}$

$\dbinom{4}{0}$   $\dbinom{4}{1}$   $\dbinom{4}{2}$   $\dbinom{4}{3}$   $\dbinom{4}{4}$

$\dbinom{5}{0}$   $\dbinom{5}{1}$   $\dbinom{5}{2}$   $\dbinom{5}{3}$   $\dbinom{5}{4}$   $\dbinom{5}{5}$

$\dbinom{6}{0}$   $\dbinom{6}{1}$   $\dbinom{6}{2}$   $\dbinom{6}{3}$   $\dbinom{6}{4}$   $\dbinom{6}{5}$   $\dbinom{6}{6}$

Take, for example,   $\dbinom{5}{3}$ and the six integers surronding it,

$\dbinom{4}{2}$   $\dbinom{4}{3}$   $\dbinom{5}{4}$   $\dbinom{6}{4}$   $\dbinom{6}{3}$   $\dbinom{5}{2}$

The product of these 6 integers is:

$\dbinom{4}{2} \cdot \dbinom{4}{3} \cdot \dbinom{5}{4} \cdot \dbinom{6}{4} \cdot \dbinom{6}{3} \cdot \dbinom{5}{2}$

$= 6 \times 4 \times 5 \times 15 \times 20 \times 10$
$= \; 360000$
$= \; 600^2$

The product of 6 integers surrounding a number interior is always a square number

$\dbinom{n}{k} \; = \; n! \,/ \,(k! \, (n-k)!)$

$\dbinom{n}{k} \; = \; \dbinom{n-1}{k-1} \; + \; \dbinom{n-1}{k}$

We have   $\dbinom{n}{k}$   surrounded by 6 integers:

……….   $\dbinom{n-1}{k-1}$   $\dbinom{n-1}{k}$

…..   $\dbinom{n}{k-1}$   $\dbinom{n}{k}$   $\dbinom{n}{k+1}$

……….   $\dbinom{n+1}{k}$   $\dbinom{n+1}{k+1}$

$\dbinom{n-1}{k-1} \cdot \dbinom{n-1}{k} \cdot \dbinom{n}{k+1} \cdot \dbinom{n+1}{k+1} \cdot \dbinom{n+1}{k} \cdot \dbinom{n}{k-1}$

$\dbinom{n-1}{k-1} \; = \; (n-1)! \,/ \,((k-1)! \, (n-k)!)$

$\dbinom{n-1}{k} \; = \; (n-1)! \,/ \,(k! \, (n-k-1)!)$

$\dbinom{n}{k+1} \; = \; n! \,/ \,((k+1)! \, (n-k-1)!)$

$\dbinom{n+1}{k+1} \; = \; (n+1)! \,/ \,((k+1)! \, (n-k)!)$

$\dbinom{n+1}{k} \; = \; (n+1)! \,/ \,(k! \, (n-k+1)!)$

$\dbinom{n}{k-1} \; = \; n! \,/ \,((k-1)! \, (n-k+1)!)$

numerator:

$(n-1)! \, (n-1)! \, n! \, (n+1)! \, (n+1)! \, n! \; = \; ( \,(n-1)! \, n! \, (n+1)! \,)^2$

denominator:

$((k-1)! \, (n-k)!) \, (k! \, (n-k-1)!) \, ((k+1)! \, (n-k-1)!) \, ((k+1)! \, (n-k)!) \, (k! \, (n-k+1)!) \, ((k-1)! \, (n-k+1)!)$

$= ((k-1)! \, k! \, (k+1)! \, (k+1)! \, k! \, (k-1)!) \, ((n-k)! \, (n-k-1)! \, (n-k-1)! \, (n-k)! \, (n-k+1)! \, (n-k+1)!)$

$= ((k-1)! \, k! \, (k+1)!)^2 \, ((n-k-1)! \, (n-k)! \, (n-k+1)!)^2$

As we can see, the product is always a square.