Sample Question 20:

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 \(N\)?

\(\textbf{(A) }9 \qquad \textbf{(B) } 14 \qquad \textbf{(C) } 16 \qquad \textbf{(D) } 17 \qquad \textbf{(E) } 19\)

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Answer Keys

Question 20: B

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Solutions

Question 20
Step 1: We know that the desired terms are in the form \(a^xb^yc^zd^w1^t\), where \(x + y + z + w + t = N\). As the 4 variables \(x\), \(y\), \(z\), and \(w\) must be at least \(1\) (whereas \(t\) can be \(0\)), let's perform a transformation by setting \(x' = x - 1\), \(y' = y - 1\), \(z' = z - 1\), and \(w' = w - 1\), thereby reducing the equation to \(x' + y' + z' + w' + t = N - 4\).

Step 2: Using stars and bars theorem (also known as ball and urn), we can see there are \(\binom{(N-4)+4}{4} = \binom{N}{4}\) ways to distribute the terms.

Step 3: We are given that there are exactly \(1001\) terms. Decomposing \(1001\) as the product of primes we get \(1001=7\cdot11\cdot13\), which provides a hint that \(N\) must be around these numbers.

Step 4: Checking the options, we find out that \(\binom{14}{4} = 1001\). So, our answer is \(\boxed{\textbf{(B) }14.}\).