find-the-contracted-form-of-n-p-2-n-p-1-n-p-2-

Question Number 74337 by Maclaurin Stickker last updated on 22/Nov/19

f i n d t h e c o n t r a c t e d f o r m o f :

(\begin{matrix} n \\ p \end{matrix}) + 2 (\begin{matrix} n \\ p + 1 \end{matrix}) + (\begin{matrix} n \\ p + 2 \end{matrix})

Answered by MJS last updated on 23/Nov/19

(\begin{matrix} n \\ p \end{matrix}) = \frac{n!}{p! (n - p)!} = \frac{n! (p + 1) (p + 2)}{p! (n - p)! (p + 1) (p + 2)}

2 (\begin{matrix} n \\ p + 1 \end{matrix}) = \frac{2 n!}{(p + 1)! (n - p - 1)!} = \frac{2 n! (n - p)}{p! (n - p)! (p + 1)} =

= \frac{2 n! (n - p) (p + 2)}{p! (n - p)! (p + 1) (p + 2)}

(\begin{matrix} n \\ p + 2 \end{matrix}) = \frac{n!}{(p + 2)! (n - p - 2)!} =

= \frac{n! (n - p) (n - p - 1)}{p! (n - p)! (p + 1) (p + 2)}

(\begin{matrix} n \\ p \end{matrix}) + 2 (\begin{matrix} n \\ p + 1 \end{matrix}) + (\begin{matrix} n \\ p + 2 \end{matrix}) =

= \frac{n! ((p + 1) (p + 2) + 2 (n - p) (p + 2) + (n - p) (n - p - 1))}{p! (n - p)! (p + 1) (p + 2)} =

= \frac{n! (n^{2} + 3 n + 2)}{p! (n - p)! (p + 1) (p + 2)} =

= \frac{n! (n + 1) (n + 2)}{p! (n - p)! (p + 1) (p + 2)} = \frac{(n + 2)!}{(p + 2)! (n - p)!} =

= \frac{(n + 2)!}{(p + 2)! ((n + 2) - (p + 2))!} = (\begin{matrix} n + 2 \\ p + 2 \end{matrix})

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