Proving factorial equations

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1. Oct 16, 2015

tesha

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number 15 questions b and c are giving me a very hard time. I have tried expanding them then factoring out the common terms but somehow not getting it to be proven. detailed help will be appreciated.

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2. Oct 16, 2015

RUber

I don't think factoring is the right answer.
Are you sure that (b) is written correctly? n choose k is equal to n choose (n-k) normally, so I don't see how the sum could be.

For (c), try writing it out and rearranging.

$\frac{n!}{k!(n-k)!} + \frac{n!}{(k-1)!(n-(k-1))!}$
And you want this to be equal to:
$\frac{n+1!}{k!((n+1)-k)!}$

**edit** You should try to make a common denominator to add the fractions. If you do this carefully and correctly, the right answer pops right out.

3. Oct 16, 2015

tesha

yes b) is written correctly

4. Oct 16, 2015

RUber

Just to be sure,
(b) says: Show that n choose k + n choose (k-1) = n choose (n-k) ?
That is straight wrong.
n choose k = n choose (n-k). This is clear from the formula factorial representation.
And n choose k + n choose (k-1) = (n+1) choose k, as in (c).

5. Oct 17, 2015

Ray Vickson

I haven't read your question (since I don't read attachments), but I assume from what RUber has written that you want to show ${}_nC_k + {}_nC_{k-1} = {}_nC_{n-k}$. As RUber has indicated, that is wrong. Try it for yourself: set n = 4, k = 2 and see what happens.

The several-century-old Passcal triangle formula says that ${}_n C _k + {n}C_{k-1} = {}_{n+1} C _k$. Again, post #2# shows you one way you can do it; another way is to start from $(1+x)^n = \sum_{k=0}^n {}_nC_k x^k$, then write $(1+x)^{n+1}$ in two ways: one way is to use the previous expansion with $n+1$ in place of $n$, and the other is to write it as $(1+x) (1+x)^n$, use the expansion for the second factor, then gather together powers of $x$.

6. Oct 18, 2015

haruspex

From the text before and after b) in the attachment, it is very clear to me that b) ought to read "show that nCk=nCn-k."

7. Oct 19, 2015

RUber

I wonder if @tesha is still working on this.
If so, then I am sure the correction haruspex suggests will help. That interpretation for (b) will also be reinforced by (d), which looks to be a direct application of that rule.