In the binomial expansion of `(3 sqrt(3) +sqrt(2))^(5)`, the term which does not contain irrational number is :

Options

A. `1^(st)`
B. `3^(rd)`
C. `4^(th)`
D. `5^(th)`

Answer

To solve the problem of finding the term in the binomial expansion of \((3\sqrt{3} + \sqrt{2})^5\) that does not contain any irrational numbers, we will follow these steps:

### Step 1: Identify the General Term
The general term \(T_{r+1}\) in the binomial expansion of \((a + b)^n\) is given by:
\[ T_{r+1} = \binom{n}{r} a^{n-r} b^r \]
In our case, \(n = 5\), \(a = 3\sqrt{3}\), and \(b = \sqrt{2}\).

### Step 2: Write the General Term
Substituting the values into the formula:
\[ T_{r+1} = \binom{5}{r} (3\sqrt{3})^{5-r} (\sqrt{2})^r \]
This can be simplified to:
\[ T_{r+1} = \binom{5}{r} \cdot 3^{5-r} \cdot (\sqrt{3})^{5-r} \cdot (\sqrt{2})^r \]
\[ = \binom{5}{r} \cdot 3^{5-r} \cdot 3^{(5-r)/2} \cdot 2^{r/2} \]
\[ = \binom{5}{r} \cdot 3^{(10-r)/2} \cdot 2^{r/2} \]

### Step 3: Determine Conditions for Rationality
For \(T_{r+1}\) to be a rational number, both exponents \((10-r)/2\) and \(r/2\) must be integers. This means:
1. \(10 - r\) must be even (i.e., \(r\) must be even).
2. \(r\) must also be even.

### Step 4: Find Possible Values of \(r\)
Since \(r\) must be even and can take values from \(0\) to \(5\) (inclusive), the possible values for \(r\) are \(0, 2, 4\).

### Step 5: Check Each Value of \(r\)
- For \(r = 0\):
\[ T_{1} = \binom{5}{0} \cdot 3^{10/2} \cdot 2^{0/2} = 1 \cdot 3^5 \cdot 1 = 243 \quad \text{(Rational)} \]

- For \(r = 2\):
\[ T_{3} = \binom{5}{2} \cdot 3^{(10-2)/2} \cdot 2^{2/2} = 10 \cdot 3^4 \cdot 2^1 = 10 \cdot 81 \cdot 2 = 1620 \quad \text{(Rational)} \]

- For \(r = 4\):
\[ T_{5} = \binom{5}{4} \cdot 3^{(10-4)/2} \cdot 2^{4/2} = 5 \cdot 3^3 \cdot 2^2 = 5 \cdot 27 \cdot 4 = 540 \quad \text{(Rational)} \]

### Conclusion
All three terms \(T_1\), \(T_3\), and \(T_5\) are rational. However, we need the term that does not contain any irrational number.

The term \(T_3\) corresponds to \(r = 2\) and is:
\[ T_3 = 1620 \]

Thus, the term which does not contain any irrational number is:
\[ \text{The answer is } 1620. \]