Question

I need help solving this without a lot of work (a-b)^9

Answer 1

ever heard of pascal's triangle?

Answer 2

Yes I have but I am not sure how to incorporate it with the problem.

Answer 3

http://mathforum.org/workshops/usi/pascal/images/pascal.hex2.gif

you use the coefficients which are on the line for 9. 1, 9 , 36, 84, ...

i'd suggest watching a youtube video to really drill this in but i'll do my best:

\[(a-b)^9 = 1a^9b^0 + 9a^8b^1 + 36a^7b^2 + 84a^6b^3 + 126a^5b^4 + 126a^4b^5 +84a^3b^6 +\]\[+36a^2b^7 + 9a^1b^8 + 1a^0b^9\]

i left the coefiicient 1 and left a^0 and b^0 to be explicit, but usually you don't need to include them.

notice that the coefficients are the numbers on pascal's triangle and the the exponent of a goes down by 1 every term while the exponent of b goes up by 1 every term.

not sure what more to say about it but feel free to ask questions ^_^

this works for any numbers a and b

Answer 4

crap i made a mistake.

i just did:\[(a+b)^9\]

while your problem is \[(a-b)^9\]

they are very similar so no worries. still worth reading ^_^

Answer 5

Thank you so much for taking your time to explain it to me.

Answer 6

Do you know what it means when they say compute \[15^C 5\]

Answer 7

for (a - b)^9, the minus sign makes the terms alternate from + to -:

so while i gave you (a+b)^9:

\[(a-b)^9 = a^9 - 9a^8b + 36a^7b^2 - 84a^6b^3 + 126a^5b^4 - 126a^4b^5 + 84a^3b^6 - 36a^2b^7 + 9ab^8 - b^9\]

i hope it all fits in one line :P

Answer 8

doesn't copy/pasting:
\[(a-b)^9 = a^9 - 9a^8b + 36a^7b^2 - 84a^6b^3 + 126a^5b^4 - 126a^4b^5\]\[+ 84a^3b^6 - 36a^2b^7 + 9ab^8 - b^9\]

Answer 9

depends which class it is for and what context.

might be that C button your calculator [probability]

Answer 10

You have been such GREAT help thank you so much!!!!! :)

Answer 11

glad i could help ^_^

Answer 12

for 15 C 5, if its a probability math class, it might be the nCr button on your calculator

Answer 13

http://www.wolframalpha.com/input/?i=expand+%28a-b%29%5E9

Answer 14

\[
{n \choose r }= \frac {n!}{r!(n-r)!}
\]