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.
 
 
 
 

 
 
 
 
 
 
 
 
 
 
 
 

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About benvitalis

math grad - Interest: Number theory
This entry was posted in Number Puzzles and tagged . Bookmark the permalink.

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