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Ok, in the Pascal Triangle, the element in the row number n and column number p is given by:
So let's take n=3 and find all the entries of that row. We are going to use 0, 1, 2 and 3 as possible values for p.
For p=0:
[tex]\frac{3!}{0!(3-0)!}=\frac{6}{1\cdot3!}=\frac{6}{6}=1[/tex]For p=1:
[tex]\frac{3!}{1!\cdot(3-1)!}=\frac{6}{2!}=\frac{6}{2}=3[/tex]For p=2:
[tex]\frac{3!}{2!\cdot(3-2)!}=\frac{6}{2\cdot1!}=\frac{6}{2}=3[/tex]And for p=3:
[tex]\frac{3!}{3!\cdot(3-3)!}=\frac{3!}{3!\cdot0!}=\frac{3!}{3!}=1[/tex]So the four entries in the third row of Pascal's Triangle are 1, 3, 3 and 1 so the statement is true.
