Difference between revisions of "2005 Alabama ARML TST Problems/Problem 12"

Revision as of 20:16, 5 January 2008

Find the number of ordered pairs of positive integers $(a,b,c,d)$ that satisfy the following equation:

$a+b+c+d=12$ .

The generating function for a is $1+x+x^2+x^3+\cdots$ . The same for b, c, and d.

$(1+x+x^2+x^3+\cdots)^4=1+4x+10x^2+\cdots$

Since the existance of the Binomial Theorem, we can assume that these are the results of choosing.

$\binom{n}{n}=1$

$\binom{n+1}{n}=4 \Rightarrow n=3$

Checking $\binom{5}{3}$ , it works.

We want the coefficient of x^12, so we have $\binom{3+12}{3}=455$ .

Revision as of 20:16, 5 January 2008 (view source) 1=2 (talk \| contribs) ← Older edit		Revision as of 20:16, 5 January 2008 (view source) 1=2 (talk \| contribs) (→‎See Also) Newer edit →
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	==See Also==		==See Also==
		+	{{ARML box\|year=2005\|state=Alabama\|num-b=11\|num-a=13}}