For some particular value of \(N\), when \((a+b+c+d+1)^N\) is expanded and like terms are combined the resulting expression contains exactly \(1001\) terms that include all four variables \(a\), \(b\), \(c\), and \(d\), each to some positive power. What is \(N\)?

amc81012 Jun 20, 2021

#1**+1 **

Idk if this is right or not, but this is my best shot at it:

Since 1 is a constant, I would treat this as a binomial:

\(((a+b+c+d)+1)^N\\=x_1\cdot(a+b+c+d)^N+x_2\cdot(a+b+c+d)^{N-1}+...x_{N}\cdot(a+b+c+d)+1\)

, where \(x_1, x_2, .... x_{N}\) are just random coefficients that are irrelevant, since we only care about the number of terms.

By the multinomial theorem (the theorem doesn't exactly state this but it will suffice for our problem), the number of terms in \((a+b+c+d)^{N}\) is equal to the number of ways to split a chunk of length N into 4 smaller nonnegative integer lengths (since there are 4 terms).

Since the problem specifies that all terms must include all variables a, b, c, and d, we can assume that the chunk N already includes at least 1 of a, b, c, and d, so we can say that the number of terms in \((a+b+c+d)^{N}\) with all variables included is equal to the number of ways to split a chunk of length N-4 into 4 smaller nonnegative integer lengths. (Notice that for N=1, 2, and 3, the value is negative, which makes sense, since\(a+b+c+d,(a+b+c+d)^2,(a+b+c+d)^3\)

all don't include any terms that include all variables.)

By stars and bars (the nonnegative version), there are exactly \({5+(N-4)-1 \choose 5-1}={N \choose 4}\), since 5 bars is needed to split the N-4 chunk to split it into 4 smaller chunks.

Notice that 14 choose 4 is 1001, so it is the answer (I won't go too in-depth because I have already written so much)

Edit: Made some adjustments to my answer, but I just noticed that there are so much better and faster answers online.

textot Jun 20, 2021