# Math Help - combinations

1. Originally Posted by Raiden_11
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

Attachment 3530
(a) is just evaluating each:

$C(7,5) = \frac {7!}{5! ( 7 - 5)!} = \frac {7!}{5!2!} = \frac {7 \cdot 6}{2} = 21$

$C(4,4) + C(5,4) + C(6,4) = \frac {4!}{4! (4 - 4)!} + \frac {5!}{4! (5 - 4)!} + \frac {6!}{4! (6 - 4)!}$

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

$= 1 + 5 + \frac {6 \cdot 5}{2}$

$= 1 + 5 + 15$

$= 21$

(b)
The second method is doing what i did for question two in your previous post, but at the end change C(5,1) to C(5,4) and C(6,2) to C(6,4), why can we do that?

C(n,r) = C(n, n - r)

so, for example, C(6,2) = C(6, 6 - 2) = C(6,4)

2. ## Thankyou!!!!!!!!!!

Thankyou To All Of Those People In Math Forum Who Heped Me In My Time Of Need.

You Have My Eternal Gratitude, I Just Wanna Know How You Guys Do So Good In Math..........is There A Secret??!!!!!!because Even I Do My Homework And I'm Having Trouble Studying For My Grade 12 Geometry Exam!!!!!!

3. Originally Posted by Raiden_11
Thankyou To All Of Those People In Math Forum Who Heped Me In My Time Of Need.

You Have My Eternal Gratitude, I Just Wanna Know How You Guys Do So Good In Math..........is There A Secret??!!!!!!because Even I Do My Homework And I'm Having Trouble Studying For My Grade 12 Geometry Exam!!!!!!
Aha! i knew there was a more efficient way to do this. we can use Pascal's formula. i think it's better for you to do (b) that way, as well as question 2 that i did earlier maybe (if you want to).

By Pascal's Formula, $C(n,r) = C(n - 1,r - 1) + C(n - 1, r)$

So, we have:

$C(7,5) = C(6,4) + C(6,5)$ .........now expand the last combination using the formula

$\Rightarrow C(7,5) = C(6,4) + C(5,4) + C(5,5)$

Now, $C(5,5) = C(4,4) = 1$

$\Rightarrow C(7,5) = C(6,4) + C(5,4) + C(4,4)$

as desired

QED

Now, wasn't that a lot nicer?!

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