If `P_1,P_2,......... P_(m+1)` are distinct prime numbers. Then the number of factors of `P_1^nP_2P_3....P_(m+1)` is :

Options

A. `m(n+1)`
B. `(n+1)2^(m)`
C. `n * 2 ^(m) `
D. `2^(nm)`

Answer

To solve the problem of finding the number of factors of the expression \( P_1^n P_2 P_3 \ldots P_{m+1} \), where \( P_1, P_2, \ldots, P_{m+1} \) are distinct prime numbers, we can follow these steps:

### Step 1: Understand the Prime Factorization
The expression \( P_1^n P_2 P_3 \ldots P_{m+1} \) is a product of prime factors. Here, \( P_1 \) is raised to the power \( n \), and each of the other primes \( P_2, P_3, \ldots, P_{m+1} \) is raised to the power of 1.

### Step 2: Identify the Exponents
In the expression:
- The exponent of \( P_1 \) is \( n \).
- The exponent of \( P_2 \) is \( 1 \).
- The exponent of \( P_3 \) is \( 1 \).
- ...
- The exponent of \( P_{m+1} \) is \( 1 \).

Thus, the exponents of the prime factors can be summarized as:
- \( P_1: n \)
- \( P_2: 1 \)
- \( P_3: 1 \)
- ...
- \( P_{m+1}: 1 \)

### Step 3: Count the Total Number of Factors
To find the total number of factors of a number given its prime factorization, we use the formula:
\[ \text{Number of factors} = (e_1 + 1)(e_2 + 1)(e_3 + 1) \ldots (e_k + 1) \]
where \( e_1, e_2, \ldots, e_k \) are the exponents of the prime factors.

In our case:
- For \( P_1 \), the exponent is \( n \), so we have \( n + 1 \).
- For \( P_2, P_3, \ldots, P_{m+1} \), each has an exponent of \( 1 \), contributing \( 1 + 1 = 2 \) for each.

Since there are \( m \) primes from \( P_2 \) to \( P_{m+1} \), we have \( m \) such terms.

### Step 4: Write the Final Expression
Thus, the total number of factors can be calculated as:
\[ \text{Number of factors} = (n + 1) \times 2^m \]

### Conclusion
The number of factors of \( P_1^n P_2 P_3 \ldots P_{m+1} \) is given by:
\[ \text{Number of factors} = (n + 1) \times 2^m \]