# Problem involving factorials and "choosing"

• Jun 8th 2011, 10:43 AM
MrT83
Problem involving factorials and "choosing"
Q: Using the identity nC{r-1} + nCr = {n+1}Cr. Find a value of n which satisfies the equation:
4C{n-1} + 4Cn = 5.

My attempted solution:

4C{n-1} + 4Cn = 5
5Cn = 5 (by the identity given)
{5!}/{n!(5-n)!}=5
4! = n!(5-n)!

Then i'm stuck - I can get the answer by trial and error using the fact that n is obviously between 0 and 4 and also by comparing with Pascal's triangle at the second line of work but I'm convinced that there must be a way by carrying on with the algebra.

Thanks in advance for any help and apologies for not using LATEX i did try but failed badly.
• Jun 8th 2011, 11:49 AM
Opalg
Quote:

Originally Posted by MrT83
Q: Using the identity \$\displaystyle {^nC_{r-1}} + {^nC_r} = {^{n+1}C_r\$. Find a value of n which satisfies the equation:
\$\displaystyle {^4C_{n-1}} + {^4C_n} = 5.\$

My attempted solution:

\$\displaystyle {^4C_{n-1}} + {^4C_n} = 5\$
\$\displaystyle {^5C_n} = 5\$ (by the identity given)
\$\displaystyle {5!}/{n!(5-n)!}=5\$
\$\displaystyle 4! = n!(5-n)!\$

Then i'm stuck - I can get the answer by trial and error using the fact that n is obviously between 0 and 4 and also by comparing with Pascal's triangle at the second line of work but I'm convinced that there must be a way by carrying on with the algebra.

Thanks in advance for any help and apologies for not using LATEX i did try but failed badly.

I think that your method is as good as any. It involves a little bit of trial and error at the end, but I think that is unavoidable here.

To see the LaTeX code, hover the cursor over the formulas. Each formula needs to be wrapped in [tex] [/tex] tags.
• Jun 8th 2011, 01:47 PM
MrT83
Quote:

Originally Posted by Opalg
To see the LaTeX code, hover the cursor over the formulas. Each formula needs to be wrapped in [tex] [/tex] tags.

Thanks! I think the problem was that I was wrapping with [tex]