How can I prove factorial equations involving difficult questions b and c?

[tesha]

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.

 

[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.

 

[tesha]

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.

yes b) is written correctly

 

[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).

 

[Ray Vickson]

yes b) is written correctly

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 $x$.

 

[haruspex]

yes b) is written correctly

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

 

[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.