Find the number of terms in the expansion of `(1+(x^(2))/(4)-x)^(5)` .

Answer

To find the number of terms in the expansion of the expression \((1 + \frac{x^2}{4} - x)^5\), we can follow these steps:

### Step 1: Identify the components of the expression
The expression is a trinomial of the form \(a + b + c\), where:
- \(a = 1\)
- \(b = \frac{x^2}{4}\)
- \(c = -x\)

### Step 2: Determine the number of terms (r)
Count the number of distinct terms in the trinomial. Here, we have:
- 1 (constant term)
- \(\frac{x^2}{4}\) (quadratic term)
- \(-x\) (linear term)

Thus, the number of terms \(r = 3\).

### Step 3: Identify the exponent (n)
The exponent to which the trinomial is raised is \(n = 5\).

### Step 4: Use the formula for the number of terms in a multinomial expansion
The formula to find the number of terms in the expansion of a multinomial \((x_1 + x_2 + ... + x_r)^n\) is given by:
\[ \text{Number of terms} = \binom{n + r - 1}{r - 1} \]
Substituting the values we have:
- \(n = 5\)
- \(r = 3\)

So, we calculate:
\[ \text{Number of terms} = \binom{5 + 3 - 1}{3 - 1} = \binom{7}{2} \]

### Step 5: Calculate \(\binom{7}{2}\)
Using the formula for combinations:
\[ \binom{n}{r} = \frac{n!}{r!(n - r)!} \]
We have:
\[ \binom{7}{2} = \frac{7!}{2!(7 - 2)!} = \frac{7 \times 6}{2 \times 1} = \frac{42}{2} = 21 \]

### Conclusion
Thus, the number of terms in the expansion of \((1 + \frac{x^2}{4} - x)^5\) is **21**.

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