Chapter 12: Q.45 (page 837)

Show that $(\begin{matrix} n \\ n - 1 \end{matrix}) = n a n d (\begin{matrix} n \\ n \end{matrix}) = 1$

Short Answer

Expert verified

If j and n are integers with $0 \leq j \leq n t h e n (\begin{matrix} n \\ j \end{matrix}) = \frac{n!}{j! (n - j)!}$

Step by step solution

Step 1. Given information

Here

(\begin{matrix} n \\ n - 1 \end{matrix}) a n d (\begin{matrix} n \\ n \end{matrix}) a r e g i v e n

we have to show that

$(\begin{matrix} n \\ n - 1 \end{matrix}) = n a n d (\begin{matrix} n \\ n \end{matrix}) = 1$

Step2. Showing nn-1=n

We know that

(\begin{matrix} n \\ j \end{matrix}) = \frac{n!}{j! (n - j)!}, j a n d n i n t e g e r s 0 \leq j \leq n t h e n (\begin{matrix} n \\ n - 1 \end{matrix}) = \frac{n!}{(n - n + 1)! (n - 1)!} = \frac{n!}{1! (n - 1)!} = \frac{n (n - 1)!}{(n - 1)!} = n

Step 3.showing nn =1

we know that

$(\begin{matrix} n \\ j \end{matrix}) = \frac{n!}{j! (n - j)!}, j a n d n i n t e g e r s 0 \leq j \leq n t h e n (\begin{matrix} n \\ n \end{matrix}) = \frac{n!}{(n - n)! (n)!} = \frac{n!}{0! \times n!} = \frac{n!}{n!} = 1$ $I f j a n d n i n t e g e r s 0 \leq j \leq n, t h e n (\begin{matrix} n \\ j \end{matrix}) = \frac{n!}{j! (n - j)!}, j a n d n i n t e g e r s 0 \leq j \leq n t h e n (\begin{matrix} n \\ n \end{matrix}) = \frac{n!}{(n - n)! (n)!} = \frac{n!}{0! (n)!} = \frac{1!}{0!} = 1$

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Show that $(\begin{matrix} n \\ n - 1 \end{matrix}) = n a n d (\begin{matrix} n \\ n \end{matrix}) = 1$

Short Answer

Step by step solution

Step 1. Given information

Step2. Showing nn-1=n

Step 3.showing nn =1

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Show thatnn-1=nandnn=1

Short Answer

Step by step solution

Step 1. Given information

Step2. Showing nn-1=n

Step 3.showing nn =1

One App. One Place for Learning.

Most popular questions from this chapter

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Show that $(\begin{matrix} n \\ n - 1 \end{matrix}) = n a n d (\begin{matrix} n \\ n \end{matrix}) = 1$