Binomial Cooeficient help check

kory

New member
Joined
Mar 8, 2021
Messages
39
Without expanding the entire expression, find the coefficient on the \(\displaystyle x^{12}y^5\) term for the binomial expansion of \(\displaystyle (2x-3y)^{17}\)

Normally I would create a tree and then simplify a long expression for these types of problems but this one says to do it without expanding.

So I'm assuming I would I do something like this?
\(\displaystyle _{17}C_5(2x)^{12}(-3y)^5\)

\(\displaystyle 17! / 12! 5!\)

\(\displaystyle 4096x^{12}(-243)y^5\)

\(\displaystyle 17 * 16 * 15* 14* 13\) / \(\displaystyle 5 * 4* 3* 2\)

\(\displaystyle 6188 * 4096 * -243 = -6,159,089,664x^{12}y^5\)

Is this still considered expanding?
 

lex

Full Member
Joined
Mar 3, 2021
Messages
480
That's perfect. The coefficient is just the number though.
(It only said without expanding the entire expression).
 

JeffM

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Sep 14, 2012
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6,328
I’d probably just write

\(\displaystyle \dbinom{17}{5} (2x)^{12}(-3y)^5 = - \dfrac{17 * 16 * 15 * 14 * 13}{5 * 4 * 3 * 2} * 2^{12} * 3^{5} x^{12}y^5 = - 17 * 13 * 7 * 3^5 * 2^{14} x^{12} y^{5}.\)

Pay attention to parentheses. With that kind of number, we might find prime factorization useful.
 

kory

New member
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Mar 8, 2021
Messages
39
Good to know. Thank you
 
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