Answered

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The n = 3 row of Pascal's Triangle has the following entries: 1, 3, 3, and 1TrueFalse

Sagot :

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.

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