# Choosing 6 letters out of 10

• September 15th 2009, 01:30 PM
Berntsson
Choosing 6 letters out of 10
I missed the lecture about this subject so now i need some help with this problem. (sry for my bad english, it ain't my nativ language).

I have 10 bricks with the letters [B],[E],[R],[N],[T],[B],[E],[R],[N],[T]
So in how many ways can i choose 6 out of this 10?

And [B][E][R][N][T][B] and [B][E][R][N][T][B] when the two b's have switcht places is counting as the same sequence of letters and shud not be counted twice. And that goes for all duplicates were 1 letters is choosen that another allredy have been on.

Whud be realy greatful for your help, i realy need to understand this to our upcomming eximination.

And if you dont know what i mean tell me and i will try to reply with a bether explanation.
• September 15th 2009, 01:44 PM
Plato
Quote:

Originally Posted by Berntsson
I have 10 bricks with the letters [B ],[E],[R],[N],[T],[B ],[E],[R],[N],[T]
So in how many ways can i choose 6 out of this 10?
And [B ][E][R][N][T][B ] and [B ][E][R][N][T][B ] when the two b's have switcht places is counting as the same sequence of letters and shud not be counted twice. And that goes for all duplicates were 1 letters is choosen that another allredy have been on.

I understand the language/translation problem.
But you have said choose six, but from the question I think that you may mean something about arrangements.
For example: [B ][E][R][N][T][B ] and [B ][E][R][N][T][B ] is only one selection.
Whereas [B ][E][R][N][T][B ] and [B ][E][R][N] [B ] [T] are two different arrangements.

• September 15th 2009, 02:25 PM
Berntsson
You are completely right, its how many arrangements of 6 letters from the 10 that you can make.
• September 15th 2009, 02:57 PM
Plato
Quote:

Originally Posted by Berntsson
I have 10 bricks with the letters [B ],[E],[R],[N],[T],[B ],[E],[R],[N],[T]
So in how many ways can I have word strings of length six?

You need to consider three cases:
1) only one letter repeated.
2) two different letters repeated.
3) three letters repeated

1) $\binom{5}{1}\left(\frac{6!}{2!}\right)$.

2) $\binom{5}{2}\binom{3}{2}\left(\frac{6!}{(2!)^2}\ri ght)$

3) $\binom{5}{3}\left(\frac{6!}{(2!)^3}\right)$
• September 15th 2009, 03:27 PM
Berntsson
Quote:

Originally Posted by Plato
You need to consider three cases:
1) only one letter repeated.
2) two different letters repeated.
3) three letters repeated

1) $\binom{5}{1}\left(\frac{6!}{2!}\right)$.

2) $\binom{5}{2}\binom{3}{2}\left(\frac{6!}{(2!)^2}\ri ght)$

3) $\binom{5}{3}\left(\frac{6!}{(2!)^3}\right)$

Do you sum them up or multiply them with eatch other after that?
• September 15th 2009, 03:52 PM
Plato
• September 16th 2009, 04:26 AM
Berntsson
Dam when i read this today i see that i was realy tierd yesterday and told you whron information. This is like picking a poker hand but you only have 10 "cards" and 5 pairs, so [B ][E][R][N][T][B ] and [B ][E][R][N] [B ] [T] is counting as the same pick and so do all the difrent arrangements with the same six letters no mather if the both T's are one in the end and one in the start. Or they are both next to eatch other.

I'm realy sorry for my mistake and bad explanation but i realy suck at english (perticular when i'm tierd).
• September 16th 2009, 05:23 AM
Plato
Quote:

Originally Posted by Berntsson
Dam when i read this today i see that i was realy tierd yesterday and told you whron information. This is like picking a poker hand but you only have 10 "cards" and 5 pairs,

In that case expand $(1+x+x^2)^5$.
One of the terms is $45x^6$.
That tells us that there 45 ways to get a hand of six cards from those ten.
• September 16th 2009, 09:15 AM
Berntsson
Quote:

Originally Posted by Plato
In that case expand $(1+x+x^2)^5$.
One of the terms is $45x^6$.
That tells us that there 45 ways to get a hand of six cards from those ten.

sorry for being slow minded but is (1+x+x^2) some general formula that you just ^with 5 becouse of the 5 pairs? Then you use the binomial theorem?
And 45 is the highest cofficient infront of x?

• September 16th 2009, 10:58 AM
Plato
Quote:

Originally Posted by Berntsson
sorry for being slow minded but is (1+x+x^2) some general formula that you just ^with 5 becouse of the 5 pairs? Then you use the binomial theorem?
And 45 is the highest cofficient infront of x?

There are entire courses on theory of generating functions. That is what I used.
Go to this website Wolfram|Alpha
In the input window, type in this exact expression: expand (1+x+x^2)^5 (you can copy & paste)
Click the equals bar at the right-hand end of the input window.

When you do the above, you will see a polynomial in powers of x from zero to ten.
The coefficient of the nth term tells us the number of ways to select n items from the ten “bricks”. As you can see that there are 51 ways to select five bricks.