# Thread: [SOLVED] Consolidated list of problems I could really use some help with!

1. ## [SOLVED] Consolidated list of problems I could really use some help with!

Im trying to learn this stuff, but it is so freaking confusing just reading a book. If I can see it done, then that helps tremendously!

1. Prove by mathematical induction that 2+5+8+...+1/2n(3n+1)

2. Find F(2), F(3), F(4), F(5) if F(n) is defined recursively by F(0)=0, f(1)=1, f(n+1)=f(n)+2f(n-1)+1 for n = 1,2,...

3. A bagel shop has onion bagels, egg bagels, sesame seed bagels and plain bagels. How many ways are there to choose twelve bagels with at least one of each kind?

4. Find the next larger permutation in lexicographic order after the permutation 42837651?

Im still rereading all the chapters trying to make sense of it. If I manage to figure them out before hand, Ill post up my answers.

2. Is this correct?
4. Find the next larger permutation in lexicographic order after the permutation 42837651?

42873651

3. Originally Posted by ninobrn99
3. A bagel shop has onion bagels, egg bagels, sesame seed bagels and plain bagels. How many ways are there to choose twelve bagels with at least one of each kind?[/FONT]

4. Find the next larger permutation in lexicographic order after the permutation 42837651?
There are $\displaystyle \binom{N+k-1}{N}$ to put N identical objects (in this case choices) into k different cells (in this case flavors of bagels) .
If we go ahead and place 2 of the objects into each of the cells, how many ways are there to deal with what remains?

Try 42851367. Does that work? WHY?

4. Originally Posted by ninobrn99
Is this correct?
4. Find the next larger permutation in lexicographic order after the permutation 42837651?

42873651
I saw you read this : http://www.mathhelpforum.com/math-he...mutations.html
As you certainly noticed, you cannot make it begin by 4283, because 7651 is already the largest possible sequence.
Then, it must begin by 428, and then be followed by the nearest (and greater) digit to 3, that is 5.
So it must begin by 4285.
Then, there are 1,3, 6 and 7 left.
Group them in increasing order and you'll get the number you want :
42851367

Looks good to you ?

5. Originally Posted by Plato
There are $\displaystyle \binom{N+k-1}{N}$ to put N identical objects (in this case choices) into k different cells (in this case flavors of bagels) .
If we go ahead and place 2 of the objects into each of the cells, how many ways are there to deal with what remains?

Try 42851367. Does that work? WHY?
It does because the next big is 5, not 7.

As for the first one, am I on the right track?
$\displaystyle \binom{12+4-1}{12}$

Originally Posted by Moo
I saw you read this : http://www.mathhelpforum.com/math-he...mutations.html
As you certainly noticed, you cannot make it begin by 4283, because 7651 is already the largest possible sequence.
Then, it must begin by 428, and then be followed by the nearest (and greater) digit to 3, that is 5.
So it must begin by 4285.
Then, there are 1,3, 6 and 7 left.
Group them in increasing order and you'll get the number you want :
42851367

Looks good to you ?
Sure does. I was just going by what I was reading in that thread by trying to flip the last two digits and when I came across my first solution, I stopped there instead of trying other combinations.

6. Originally Posted by ninobrn99
It does because the next big is 5, not 7.

As for the first one, am I on the right track?
$\displaystyle \binom{12+4-1}{12}$
No, not quite.
It is $\displaystyle \binom{{\color{red}8}+4-1}{{\color{red}8}}$ because you put one choice into each flavor leaving only 8 more choices.

7. Originally Posted by Plato
No, not quite.
It is $\displaystyle \binom{{\color{red}8}+4-1}{{\color{red}8}}$ because you put one choice into each flavor leaving only 8 more choices.
Sorry for being slow on it...im trying!

Okay, so then:

$\displaystyle \binom{{\color{red}8}+4-1}{{\color{red}8}}$
c=(11,8)

11*10*9*8*7*6*5*4 / 1*2*3*4
=6652800 / 24
=277200

8. $\displaystyle \binom{11}{8} =\frac{11!}{(8!)(3!)} = \frac{11\cdot 10\cdot 9}{3\cdot 2\cdot 1}= 165$

9. Originally Posted by Plato
$\displaystyle \binom{11}{8} =\frac{11!}{(8!)(3!)} = \frac{11\cdot 10\cdot 9}{3\cdot 2\cdot 1}= 165$
Where did the 3 Fitario come from? Thank you for the patience and assitance!!

10. Originally Posted by ninobrn99
Where did the 3 Fitario come from?
It is your job and responsibility to learn the formulas.
$\displaystyle \binom{N}{k} = \frac{N!}{(k!)(N-k)!}$

11. point taken. thanks again.