A nine - digit number is formed using the digits 1, 2, 3, 5 and 7. The probability that the product of all digits is always 1920 is

Options

A. `(1)/(5^(9))`
B. `(7)/(5^(8))`
C. `(72)/(5^(9))`
D. `(1)/(7!)`

Answer

To solve the problem, we need to find the probability that the product of the digits of a nine-digit number formed using the digits 1, 2, 3, 5, and 7 equals 1920.

### Step-by-Step Solution:

**Step 1: Calculate the Total Number of Nine-Digit Combinations**

Since we are forming a nine-digit number using the digits 1, 2, 3, 5, and 7, and we can repeat the digits, the total number of combinations can be calculated as follows:

- For each of the 9 positions, we have 5 choices (1, 2, 3, 5, or 7).

Thus, the total number of nine-digit combinations is:
\[ \text{Total Cases} = 5^9 \]

**Step 2: Prime Factorization of 1920**

Next, we need to find the prime factorization of 1920 to understand how we can form the product using the digits.

- Dividing 1920 by 2 repeatedly:
- \(1920 \div 2 = 960\)
- \(960 \div 2 = 480\)
- \(480 \div 2 = 240\)
- \(240 \div 2 = 120\)
- \(120 \div 2 = 60\)
- \(60 \div 2 = 30\)
- \(30 \div 2 = 15\)

- Now, 15 is not divisible by 2, but it is divisible by 3 and 5:
- \(15 \div 3 = 5\)
- \(5 \div 5 = 1\)

Thus, the prime factorization of 1920 is:
\[ 1920 = 2^7 \times 3^1 \times 5^1 \]

**Step 3: Determine the Favorable Cases**

We need to form the number such that the product of the digits equals \(1920\). Given the digits available (1, 2, 3, 5, 7), we can only use 2, 3, and 5 from our digits to achieve the required product.

- We need:
- 2 appears 7 times
- 3 appears 1 time
- 5 appears 1 time

Since we have 9 digits in total, we can fill the remaining 6 positions with the digit 2, and then place 3 and 5 in the remaining positions.

**Step 4: Arranging the Digits**

The arrangement of the digits can be calculated using the formula for permutations of multiset:
\[ \text{Favorable Cases} = \frac{9!}{7! \times 1! \times 1!} = \frac{9 \times 8}{1 \times 1} = 72 \]

**Step 5: Calculate the Probability**

Finally, we can find the probability that the product of the digits equals 1920:
\[ \text{Probability} = \frac{\text{Favorable Cases}}{\text{Total Cases}} = \frac{72}{5^9} \]

### Final Answer

Thus, the probability that the product of all digits is 1920 is:
\[ \frac{72}{5^9} \]