Question

The number of terms in the expansion of \( (2x + 3y + 5z)^5 \) is

• A. 18
• B. 6
• C. 21
• D. 15

Hint:

The number of terms in the multinomial expansion of \( (a + b + c)^n \) is given by \( \binom{n + k - 1}{k - 1} \), where \( k \) is the number of variables.

Answer 1

The Correct Option is C

Step 1: Identify the multinomial expansion.
In the expansion of \( (2x + 3y + 5z)^5 \), the number of terms corresponds to the number of distinct combinations of powers of \( x, y, z \) in the expansion. Each term in the expansion is of the form \( (2x)^a (3y)^b (5z)^c \), where \( a + b + c = 5 \) and \( a, b, c \geq 0 \).
Step 2: Apply the stars and bars method.
The number of distinct non-negative integer solutions to the equation \( a + b + c = 5 \) is given by the formula for combinations with repetition: \[ \binom{5 + 3 - 1}{3 - 1} = \binom{7}{2} = 21. \]
Step 3: Conclusion.
Thus, the number of terms in the expansion is 21, which corresponds to option (C).

Answer 2

Instead of applying the stars-and-bars formula immediately, let's build up the count by reasoning about how many values the exponent of \( z \) can take, and for each such value, how many terms remain, then sum over all cases.

In each term \( (2x)^a(3y)^b(5z)^c \) with \( a+b+c=5 \), fix \( c \) (the power of \( z \)) at each possible value from \( 0 \) to \( 5 \). For a fixed \( c \), we need \( a+b = 5-c \) with \( a,b\ge0 \), which has \( 5-c+1 = 6-c \) solutions. Summing over \( c=0,1,\dots,5 \): \[ \sum_{c=0}^{5}(6-c) = 6+5+4+3+2+1 = 21. \]

1. 18: This does not match the sum \( 6+5+4+3+2+1 \) obtained by fixing each power of \( z \) in turn, so it is incorrect.
2. 6: This is just the number of values \( c \) can take (0 through 5), not the total count of terms, so it is incorrect.
3. 21: This matches the summed count exactly.
4. 15: This is the count of two-variable expansions like \( (x+y)^5 \) (which has \( 6 \) terms) or a related smaller combinatorial count; it doesn't match this three-variable case and is incorrect.

Summing the term counts over each fixed power of \( z \) reproduces the stars-and-bars total.

Therefore, the correct answer is 21.