The number of ways of selecting ' n ' things out of '3n ' things of which 'n ' are of one kind and alike and 'n ' are of second kind and alike and the rest unlike is :

Options

A. `n2^(n-1)`
B. `(n-1)2^(n-1)`
C. `(n+1)2^(n-1)`
D. `(n+2)2^(n-1)`

Answer

To solve the problem of selecting 'n' things out of '3n' things where 'n' are of one kind and alike, 'n' are of a second kind and alike, and the remaining 'n' are all unlike, we can follow these steps:

### Step 1: Understand the Composition of the Objects
We have:
- n objects of the first kind (alike)
- n objects of the second kind (alike)
- n objects of the third kind (all unlike)

### Step 2: Define the Selection Process
We need to select 'n' objects from these '3n' objects. The selection can be broken down based on how many objects we choose from the first two kinds (which are alike).

Let:
- \( k_1 \) = number of objects chosen from the first kind
- \( k_2 \) = number of objects chosen from the second kind
- \( k_3 \) = number of objects chosen from the third kind

We have the equation:
\[ k_1 + k_2 + k_3 = n \]
where \( k_1 \) can take values from 0 to n, and \( k_2 \) can also take values from 0 to n.

### Step 3: Count the Choices
For each possible value of \( k_1 \) (from 0 to n), we can determine \( k_2 \) and \( k_3 \) as follows:
- If we choose \( k_1 \) from the first kind, then we can choose \( k_2 \) from the second kind, and the remaining objects will be chosen from the third kind.

The number of ways to choose \( k_1 \) from the first kind is 1 (since they are alike), the number of ways to choose \( k_2 \) from the second kind is also 1, and the number of ways to choose \( k_3 \) from the third kind is given by \( \binom{n}{k_3} \).

### Step 4: Summing Over All Possible Choices
We need to sum over all possible values of \( k_1 \) and \( k_2 \). The total number of ways can be expressed as:
\[ \sum_{k_1=0}^{n} \sum_{k_2=0}^{n} \binom{n}{n - k_1 - k_2} \]

### Step 5: Using the Binomial Theorem
Using the binomial theorem, we can simplify the sum:
\[ \sum_{k=0}^{n} \binom{n}{k} = 2^n \]

### Step 6: Final Calculation
The total number of ways to select 'n' objects from '3n' objects is given by:
\[ (n + 1) \cdot 2^{n - 1} \]

### Final Answer
Thus, the number of ways of selecting 'n' things out of '3n' things is:
\[ (n + 2) \cdot 2^{n - 1} \]