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**cooljackiec****Member**- Registered: 2012-12-13
- Posts: 160

how can we prove 2n choose n is always even?

I see you have graph paper.

You must be plotting something

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,385

hi cooljackiec

Welcome to the forum.

I'm not at all clear what you are asking. Please would you say a bit more about this problem.

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,476

Hi;

I like this proof best:

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**cooljackiec****Member**- Registered: 2012-12-13
- Posts: 160

so because i have a 2 * 2n-1 choose n, it is bound to be even?

I see you have graph paper.

You must be plotting something

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,476

Hi;

Yes, any integer that is multipled by 2 is even.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**cooljackiec****Member**- Registered: 2012-12-13
- Posts: 160

Another problem:

Consider the polynomial

What are the coefficients of $ f(t-1) $? Enter your answer as an ordered list of four numbers. For example, if your answer were $ f(t-1) = t^3+3t^2-2t+7 $, you'd enter (1,3,-2,7). (This is not the actual answer.)

I see you have graph paper.

You must be plotting something

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,476

Hi;

I am getting (1, 3, 3, 1).

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**cooljackiec****Member**- Registered: 2012-12-13
- Posts: 160

thank you. I have another question:

We have 8 pieces of strawberry candy and 7 pieces of pineapple candy. In how many ways can we distribute this candy to 4 kids?

and:

In how many ways can we distribute 13 pieces of identical candy to 5 kids, if the two youngest kids are twins and insist on receiving an equal number of pieces?

and:

4 students are running for club president in a club with 50 members. How many different vote counts are possible, if members may choose not to vote?

I see you have graph paper.

You must be plotting something

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**noelevans****Member**- Registered: 2012-07-20
- Posts: 236

I'm getting the same as bobbym. It sometimes helps to write the original function in terms of

a different variable than the one you are substituting in its place. For example

f(x) = x^3+6x^2+12x+8 = x^3 + 2*3x^2 + 4*3x + 8 = (x+2)^3

f(x) = (x+2)^3 so f(t-1) = ((t-1)+2)^3 = (t+1)^3 = t^3+3t^2+3t+1 = (1, 3, 3, 1) by replacing

x by t-1 in the f(x) = (x+2)^3.

And as another example find the quadruple for f(t-2).

f(x) = (x+2)^3 so f(t-2) = ((t-2)+2)^3 = t^3 = (3, 0, 0, 0).

These can also be looked at as a composition of functions: f(x)=(x+2)^3 and g(x)=x-1.

(fog)(x) = f(g(x)) = f(x-1) = ( (x-1)+2)^3 = (x+1)^3

Writing "pretty" math (two dimensional) is easier to read and grasp than LaTex (one dimensional).

LaTex is like painting on many strips of paper and then stacking them to see what picture they make.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,476

Hi cooljackiec;

4 students are running for club president in a club with 50 members. How many different vote counts are possible, if members may choose not to vote?

I am getting 316251.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,522

On the other hand, I am getting 5^50.

Here lies the reader who will never open this book. He is forever dead.

Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,476

5^50 = 88817841970012523233890533447265625

I bet you did not do a simulation.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,522

Each member has 5 chiloices. There are 50 members, or 46 if you exclude the ones who are running for president. 5 votes per person, 46 persons, 5^46 possible vote counts...

Here lies the reader who will never open this book. He is forever dead.

Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,476

Hi;

Not exactly, If one guy gets 20 votes there are only 30 to spread to the others.

Remember you are only voting for one position not 4.

Take 50 x's and place 3 spacers in various positions.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,522

Actually, you must have 4 spacers...

Your answer is correct.

Here lies the reader who will never open this book. He is forever dead.

Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,476

3 spacers, because you are looking for solutions to

the three spacers make 4 separte groups. Each group is how many is in a variable.

xxxxxxxxxxxxx _ xxxxxxxxxxxxxxxxx _ xxxxxxxxxx _ xxxxxxxxxx

this corresponds to the solution 13 + 17 + 10 + 10 = 50

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,522

There are 5 groups, votes for 1st, 2nd, 3rd and 4th candidate and the non-voters...

Here lies the reader who will never open this book. He is forever dead.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,476

The non voters are not a candidate. They are represented by different values of r. For instance when there is one non voter the equation is

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,522

It is easier to look at them as a special category of voters:

where n are the non-voters. There are then

.Here lies the reader who will never open this book. He is forever dead.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,476

Hi;

That is very good, I did not see that. That would get the same answer and there is less calculation.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,522

Exactly!

Here lies the reader who will never open this book. He is forever dead.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,476

Exactly?

There will be times when you will think you or I have found the perfect answer, I assure you these are delusions on your part. - Prof. Kingsfield

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,522

I never said it was perfect... I only agreed that it is much easier to calculate...

Here lies the reader who will never open this book. He is forever dead.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,476

True, but there could be an even easier way...

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**cooljackiec****Member**- Registered: 2012-12-13
- Posts: 160

it is wrong.

I think that it would be 4^50.

Every member has 4 choices. 50 members. But im not sure...

I see you have graph paper.

You must be plotting something

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