nCr 'show that' qu.

**WWTL@WHL** · November 12th 2007, 10:43 AM

Questions:

a) Suppose that r and n are integers with $1 \le r \le n$ .

Show that $^n\mathrm{C}_r + ^n\mathrm{C}_{r-1} = ^{n+1}\mathrm{C}_r$

b) If n is a positive integer,

Show that $2^n= ^n\mathrm{C}_0 + ^n\mathrm{C}_1 + ^n\mathrm{C}_2 +......+ ^n\mathrm{C}_{n-1} + ^n\mathrm{C}_n$

__________________________________________________ _____________

My attempt so far:

a) $^n\mathrm{C}_r + ^n\mathrm{C}_{r-1} = \frac{n!}{(n-r)!r!} + \frac{n!}{(n-(r-1))!(r-1)!}$

$^n\mathrm{C}_r + ^n\mathrm{C}_{r-1} = \frac{n! (n-r+1)!(r-1)! + n!(n-r)!r!}{(n-r)!r!(n-r+1)!(r-1)!}$

Ugh! Any ideas where I could go from here?

b) $2^n = \sum (^n\mathrm{C}_n) = \sum \frac{n!}{(n-n)!n!} = \sum \frac{n!}{n!} = \sum 1$

which is clearly wrong.

Please help!

Thank you!

**Plato** · November 12th 2007, 11:44 AM

$\begin{array}{rcl} \frac{{n!}}{{\left( {n - r} \right)!r!}} + \frac{{n!}}{{\left( {n + 1 - r} \right)!\left( {r - 1} \right)!}} & = & \frac{{n!\left( {n + 1 - r} \right) + n!r}}{{\left( {n + 1 - r} \right)!r!}} \\ & = & \frac{{n!\left( {n + 1} \right) - n!r}}{{\left( {n + 1 - r} \right)!r!}} \\ & = & _{n + 1} C_r \\ \end{array} $

For the next one.
$\left( {x + y} \right)^n = \sum\limits_{}^{} {_n C_k x^k y^{n - k} }$
Let x=1 & y=1.

**WWTL@WHL** · November 12th 2007, 02:45 PM

Originally Posted by Plato

$\begin{array}{rcl} \frac{{n!}}{{\left( {n - r} \right)!r!}} + \frac{{n!}}{{\left( {n + 1 - r} \right)!\left( {r - 1} \right)!}} & = & \frac{{n!\left( {n + 1 - r} \right) + n!r}}{{\left( {n + 1 - r} \right)!r!}} \\ & = & \frac{{n!\left( {n + 1} \right) - n!r}}{{\left( {n + 1 - r} \right)!r!}} \\ & = & _{n + 1} C_r \\ \end{array} $

For the next one.
$\left( {x + y} \right)^n = \sum\limits_{}^{} {_n C_k x^k y^{n - k} }$
Let x=1 & y=1.

For the first part, that's brilliant! I'll make a note of that trick. It's a lot easier than the way I tried to do it.

And for (b), subtituting x=1, y=1 gives

$2^n = \sum nC_k$

How does the next step follow?

Thanks for the help, once again!

**Plato** · November 12th 2007, 03:25 PM

Originally Posted by WWTL@WHL

How does the next step follow?

Oh! Come on! EXPAND.
$\sum\limits_{k = 0}^N {_N C_k } = _N C_0 + _N C_1 + \cdots + _N C_N$

**WWTL@WHL** · November 12th 2007, 03:29 PM

Originally Posted by Plato

Oh! Come on! EXPAND.
$\sum\limits_{k = 0}^N {_N C_k } = _N C_0 + _N C_1 + \cdots + _N C_N$

What?! In my textbook it says the sum of nCr goes to $_n C_1 + _n C_2+.....+ _N C_ {r-1} + _N C_r$

I'll have to check somewhere else. I used to be able to do this stuff without a second thought 2 years ago. It's been so long since I've touched stats.

But thanks, Plato. You're a life-saver!

**Plato** · November 12th 2007, 03:41 PM

Originally Posted by WWTL@WHL

Show that $2^n= ^n\mathrm{C}_0 + ^n\mathrm{C}_1 + ^n\mathrm{C}_2 +......+ ^n\mathrm{C}_{n-1} + ^n\mathrm{C}_n$

That is exactly what I just posted.
The text agrees.

**WWTL@WHL** · November 12th 2007, 03:50 PM

Originally Posted by Plato

That is exactly what I just posted.
The text agrees.

No, I get that. It's just the $\sum n C k$ goes to nC1 + nC2 +....+ nCn bit I didn't understand - if you understand what I'm saying.

:

All is well now, I had to look it up in one of my old high-school textbooks, and I get it now.

Sorry for the confusion on the last bit. You were extremely helpful, yet again. Thanks.

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