Show that $\displaystyle C(8,4)=C(7,3)+C(7,4)$
Let's manipulate the left side to become the right side
$\displaystyle \frac{8!}{4!\cdot (8-4)!}=\frac{8!}{4!\cdot 4!}=\frac{8\cdot 7!}{4!\cdot 4\cdot 3!}=\frac{2\cdot 7!}{4!\cdot 3!}=\frac{7!}{4!\cdot 3!}+\frac{7!}{4!\cdot 3!}$
$\displaystyle \frac{7!}{3!\cdot (7-3)!}+\frac{7!}{4!\cdot (7-4)!}=\boxed{C(7,3)+C(7,4)}$ which was to be shown.
you do if the professor decides to make your homework grade a part of your final grade
4) $\displaystyle \sum_{i = 1}^{5} i \left[ \sum_{j = 1}^{3} (2j + 1) \right] = (1 + 2 + 3 + 4 + 5) \left[ (2(1) + 1) + (2(2) + 1) + (2(3) + 1) \right]$
$\displaystyle = 15 ( 3 + 5 + 7)$
$\displaystyle = 15 \cdot 15$
$\displaystyle = 225$
#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:
$\displaystyle \sum_{j = 3}^{6} (2j - 2) = \sum_{j = 1}^{6} (2j - 2) - \sum_{j = 1}^{2} (2j - 2)$
$\displaystyle = \sum_{j = 1}^{4} (2j - 2) + \sum_{j = 5}^{6} (2j - 2) - \sum_{j = 1}^{2} (2j - 1)$ .......evaluating the last two summations we obtain
$\displaystyle = \sum_{j = 1}^{4} (2j - 2) + 16$ ........now let's do some manipulation on the first summation
$\displaystyle = \sum_{j = 1}^{4}(2j + 2 - 4) + 16$ .......i didn't change anything, +2 - 4 = -2
$\displaystyle = \sum_{j = 1}^{4} (2j + 2) + \sum_{j = 1}^{4} (-4) + 16$ ......now evaluate the second summation
$\displaystyle = \sum_{j = 1}^{4} (2j + 2) - 16 + 16$
$\displaystyle = \sum_{j = 1}^{4} (2j + 2)$ ........a bit more manipulation on this summation should do it
$\displaystyle = \sum_{j = 1}^{4} (2j + 4 - 2)$ .........again, I changed nothing, +4 - 2 = +2
$\displaystyle = \sum_{j = 1}^{4} \left[ 2(j + 2) - 2 \right]$ .........i factored out a 2 from the first two terms
and now we have the desired summation
QED
Did you understand?
here's one way to do it:
$\displaystyle C(7,5) = \frac {7!}{5! ( 7 - 5)!}$
$\displaystyle = \frac {7!}{5! 2!}$
$\displaystyle = \frac {7 \cdot 6!}{5! 2!}$
$\displaystyle = \frac {(2 + 5) \cdot 6!}{5! 2!}$
$\displaystyle = \frac {2 \cdot 6! + 5 \cdot 6!}{5! 2!}$
$\displaystyle = \frac {2 \cdot 6!}{5! 2!} + \frac {5 \cdot 6!}{5! 2!}$
$\displaystyle = 6 + \frac {6!}{4! 2!}$
$\displaystyle = 1 + 5 + \frac {6!}{4! 2!}$
$\displaystyle = C(4,4) + C(5,1) + C(6,2)$
QED
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