Binomial (1 Viewer)

[OLDMAN]

Some of the following have been discussed already. It might be useful to bring them all together to show that there usually is physical meaning behind the coefficient.

Use a combinatorial argument to show :

a) nCr = nC n-r

b) nCr = n-1 C r-1 + n-1 Cr

c) (nCo)^2 + (nC1)^2 + (nC2)^2+...+(nCn)^2 = 2n Cn

d) Sum(0-->k) [ (nCj)(mC k-j)] = n+m Ck , m>=k

 

[:: ck ::]

no idea wot combinatoral argument means .. we havent done binomial at skool either so yeh.. im taking a guess with wut i kno but its prolli wrong

a)
LHS = nCr = (n!)/[(n-r)!r!]
RHS = nC(n-r) = n! / [(n-(n-r))! (n-r)!]
= n!/[(n-r)!r!]

is this the rite way to do it? .. if it is ill attempt the others too

 

[OLDMAN]

Originally posted by :: ryan.cck ::
no idea wot combinatoral argument means .. we havent done binomial at skool either so yeh.. im taking a guess with wut i kno but its prolli wrong

a)
LHS = nCr = (n!)/[(n-r)!r!]
RHS = nC(n-r) = n! / [(n-(n-r))! (n-r)!]
= n!/[(n-r)!r!]

is this the rite way to do it? .. if it is ill attempt the others too

The algebra is right anyway, by all means practise on the rest.

But a combinatorial argument uses the definition of
nCr as the number of ways of choosing r objects from n distinct objects or "n choose r" for short.

 

[Heinz]

b) nCr = n-1 C r-1 + n-1 Cr

LHS = n!/(n-r)!(r)!
RHS = (n-1)!/(n-1-r+1)!(r-1)! + (n-1)!/(n-1-r)!(r)!
= (n-1)!/(n-r)(n-r-1)!(r-1)! + (n-1)!/(n-1-r)!(r)(r-1)!
= [(n-1)!/(n-r-1)!(r-1)!][1/n-r + 1/r]
= [(n-1)!/(n-r-1)!(r-1)!][(r)/r(n-r) + (n-r)/r(n-r)]
= [(n-1)!/(n-r-1)!(r-1)!][n/r(n-r)]
= n!/(n-r)!(r)!

c and d look hard...

 

[maniacguy]

He meant using the physical meaning behind the algebra.

For instance:
a)
nCr = number of ways to choose r objects to include in a set from a given collection of n objects
= number of ways to choose n-r objects not to include in that set from the given collection of n objects
= nC(n-r)

It's not a difficult exercise to go through the others and translate them into this form. It develops your mathematical intuition...

(notice also that (c) is a special case of (d)))

 

[CM_Tutor]

(c) can also be done by appropriately exapanding the identity (1 + x)²ⁿ = xⁿ(1 + xⁿ)(1 + 1 / xⁿ), or by alternately expanding (1 + x)²ⁿ = (1 + x)ⁿ(1 + x)ⁿ, and then applying the symmetry property (a).

 

[OLDMAN]

c) (nCo)^2 + (nC1)^2 + (nC2)^2+...+(nCn)^2 = 2n Cn
____________________________________________________
ryan.cck:
no idea wot combinatoral argument means .. we havent done binomial at skool either so yeh..
____________________________________________________

c) could be done by applying counting principles in your "permutation and combination" topic, usually discussed at the start of Yr11 Ext1 to frighten off or reduce the number of students in the Extension class (please, I am just joking!)

Here's a primer on combination, a retake of Grey Council's argument.

No. of ways of choosing n from 2n objects is 2n C(hoose)n = RHS

This could also be done by splitting the 2n objects into 2 sets of n objects:

Choose 0 objects in 1st set and n objects from 2nd set, or 1 object in 1st and n-1 in 2nd, or 2 in 1st and n-2 in 2nd, or etc. up to n objects in 1st and 0 in the 2nd.

From counting principles : "and" means "times" and "or" means "plus"

Thus nCo*nCn+nC1*nCn-1 + nC2*nCn-2 +...+nCn*nCo =
(nCo)^2 + (nC1)^2 + (nC2)^2+...+(nCn)^2 =LHS using a).

Yes, could also be done using CM_tutor's way.

maniac is right about c) being a special case of d).

btw d) has got a fancy name : Vandermonde's Convolution Formula

 

[sammeh]

usually discussed at the start of Yr11 Ext1 to frighten off or reduce the number of students in the Extension class (please, I am just joking!)

well it reduced the size of our 3u class by about 50% in a space of half a term...joking? oh.

as for the question, is a good example of the 4u syllabus outcome concerning the appreciation of creative application of maths to problem solving. not hard maths as such, but just a different way of thinking about it

 

[OLDMAN]

a fellow mudgeeite! Haven't met but heard a lot about you. Hope you're fully recovered.

 

[sammeh]

heard a lot about me? wow... a minor status of celebrity, im known by reputation! lol anyway i think we may have met in passing on a couple of occasions.

back to catching up my 5 weeks missed work...i'd LIKE to think im fully recovered but well...wish me luck!

 

[OLDMAN]

____________________________________________________
Sammeh:
heard a lot about me?
____________________________________________________

Don't worry sammeh, its just American politese when an introduction is made.

 

[sammeh]

*whew* well thats ok then.

 

[Grey Council]

wouldn't it be like 2^8 times 6463 (whatever the figure was, just deleted it from my calculator)

 

[Grey Council]

pardon? I did use the standard HSC method. Pascal's triangle.

I got the 15th row, guessed the eigth one would have highest coeffecient. . . o shit

shit shit shit

it doesn't have to, does it?
>.<

 

[Archman]

well after using the "standard hsc method" the answer is 15C10 * 2^10. Unless you want to argue that the expansion is not a polymonial and the term "coefficients" only apply to to terms with postivie integer degree..

 

[CM_Tutor]

Find the greatest coeffcient in the expansion of (x-2/x)^15

Try using the standard HSC method for this one....something goes wrong

I got that the greatest coefficient was 2¹⁰*3*7*11*13 = 3075072, by the standard HSC method. Expanding using a computer, this seems to be correct, so I'm not following what goes wrong.?

 

[Archman]

hmmm thats whay i did, whats wrong with it?

T_k+1/T_k
= 15C(k+1)*2^(15-k-1)/15Ck*2^(15-k)
= (15-k)/(k+1)*2
now its > 1 for k = 1 to 4 and < 1 for others
so the maximum occurs at k = 5

 

[maniacguy]

Perhaps our esteemed friend is attempting to distinguish between maximum coefficient and coefficient of greatest magnitude?

 

[turtle_2468]

dodgy defns... I know all about it... lol

 

[Archman]

again there is nothing wrong with my soln, it is the largest in both senses to the original proposed question.

Originally posted by turtle_2468
dodgy defns... I know all about it... lol

haha yes i saw that coming.