# Math Help - combinations

1. ## combinations

2. Why does that look like an online test?

3. Show that $C(8,4)=C(7,3)+C(7,4)$

Let's manipulate the left side to become the right side
$\frac{8!}{4!\cdot (8-4)!}=\frac{8!}{4!\cdot 4!}=\frac{8\cdot 7!}{4!\cdot 4\cdot 3!}=\frac{2\cdot 7!}{4!\cdot 3!}=\frac{7!}{4!\cdot 3!}+\frac{7!}{4!\cdot 3!}$
$\frac{7!}{3!\cdot (7-3)!}+\frac{7!}{4!\cdot (7-4)!}=\boxed{C(7,3)+C(7,4)}$ which was to be shown.

4. Originally Posted by janvdl
Why does that look like an online test?
It probably is. Or online homework anyway.

-Dan

5. Originally Posted by topsquark
It probably is. Or online homework anyway.

-Dan
Usually you dont get points for homework...

6. Originally Posted by janvdl
Usually you dont get points for homework...

4) $\sum_{i = 1}^{5} i \left[ \sum_{j = 1}^{3} (2j + 1) \right] = (1 + 2 + 3 + 4 + 5) \left[ (2(1) + 1) + (2(2) + 1) + (2(3) + 1) \right]$

$= 15 ( 3 + 5 + 7)$

$= 15 \cdot 15$

$= 225$

7. ## What about the other ones?

8. Originally Posted by Raiden_11
#2 is very similar to #1 which rualin so graciously did for you. try it on your own and tell us what you come up with.

for problem #3, i'd do something like this. chances are there's a more efficient way, but this is how i saw the solution:

$\sum_{j = 3}^{6} (2j - 2) = \sum_{j = 1}^{6} (2j - 2) - \sum_{j = 1}^{2} (2j - 2)$

$= \sum_{j = 1}^{4} (2j - 2) + \sum_{j = 5}^{6} (2j - 2) - \sum_{j = 1}^{2} (2j - 1)$ .......evaluating the last two summations we obtain

$= \sum_{j = 1}^{4} (2j - 2) + 16$ ........now let's do some manipulation on the first summation

$= \sum_{j = 1}^{4}(2j + 2 - 4) + 16$ .......i didn't change anything, +2 - 4 = -2

$= \sum_{j = 1}^{4} (2j + 2) + \sum_{j = 1}^{4} (-4) + 16$ ......now evaluate the second summation

$= \sum_{j = 1}^{4} (2j + 2) - 16 + 16$

$= \sum_{j = 1}^{4} (2j + 2)$ ........a bit more manipulation on this summation should do it

$= \sum_{j = 1}^{4} (2j + 4 - 2)$ .........again, I changed nothing, +4 - 2 = +2

$= \sum_{j = 1}^{4} \left[ 2(j + 2) - 2 \right]$ .........i factored out a 2 from the first two terms

and now we have the desired summation

QED

Did you understand?

9. ## Are u sure?

Are u sure there aren't any more steps?

10. ## I still don't know?

I still don't know how to do #2.

11. Originally Posted by Raiden_11
Are u sure there aren't any more steps?
more steps for what?

you should click the "quote" button when you're responding to something, so people know exactly what you're responding to

12. ## What i mean is....

Originally Posted by Jhevon
more steps for what?

you should click the "quote" button when you're responding to something, so people know exactly what you're responding to

Are you sure there aren't anymore steps for number 3....
also i still don't know how to do number 2..

13. Originally Posted by Raiden_11
I still don't know how to do #2.
here's one way to do it:

$C(7,5) = \frac {7!}{5! ( 7 - 5)!}$

$= \frac {7!}{5! 2!}$

$= \frac {7 \cdot 6!}{5! 2!}$

$= \frac {(2 + 5) \cdot 6!}{5! 2!}$

$= \frac {2 \cdot 6! + 5 \cdot 6!}{5! 2!}$

$= \frac {2 \cdot 6!}{5! 2!} + \frac {5 \cdot 6!}{5! 2!}$

$= 6 + \frac {6!}{4! 2!}$

$= 1 + 5 + \frac {6!}{4! 2!}$

$= C(4,4) + C(5,1) + C(6,2)$

QED

14. Originally Posted by Raiden_11
Are you sure there aren't anymore steps for number 3....
also i still don't know how to do number 2..
no, number 3 is complete. i started with one series and manipulated it and ended up with the other series. this proves the claim that they are equal. in fact, i think i probably did too many steps, i'm not very efficient with this sort of thing.

15. ## This is my final question for math forum!!

This is my final question for math forum which i hope you can answer!

1. On page 345 of the textbook, the diagonal pattern of Pascal’s Triangle illustrates that C(7, 5) = C(4, 4) + C(5, 4) + C(6, 4). Prove this relationship using the following methods.

a)numerically by using factorials
b)by reasoning, using the meaning of combinations

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