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A nine - digit number is formed using th...

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

A

`(1)/(5^(9))`

B

`(7)/(5^(8))`

C

`(72)/(5^(9))`

D

`(1)/(7!)`

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem of finding the probability that the product of all digits in a nine-digit number formed using the digits 1, 2, 3, 5, and 7 equals 1920, we can follow these steps: ### Step 1: Understand the Problem We need to form a nine-digit number using the digits 1, 2, 3, 5, and 7. The product of these digits must equal 1920. ### Step 2: Total Number of Cases Since we can use each digit multiple times, for each of the 9 positions in the number, we have 5 choices (1, 2, 3, 5, or 7). Therefore, the total number of possible nine-digit combinations is: \[ \text{Total combinations} = 5^9 \] ### Step 3: Prime Factorization of 1920 Next, we need to find the prime factorization of 1920 to understand which digits can be used to achieve this product. \[ 1920 = 2^7 \times 3^1 \times 5^1 \] ### Step 4: Determine the Digits Needed From the prime factorization, we see that: - The digit '2' can contribute to the factor of \(2^7\). - The digit '3' contributes to \(3^1\). - The digit '5' contributes to \(5^1\). ### Step 5: Count the Required Digits To achieve the product of 1920, we need: - Seven '2's - One '3' - One '5' This gives us a total of 9 digits (7 + 1 + 1 = 9), which fits our requirement for a nine-digit number. ### Step 6: Calculate the Arrangements The number of ways to arrange these digits (7 '2's, 1 '3', and 1 '5') can be calculated using the formula for permutations of multiset: \[ \text{Arrangements} = \frac{9!}{7! \times 1! \times 1!} \] Calculating this: \[ 9! = 362880 \quad \text{and} \quad 7! = 5040 \] Thus, \[ \text{Arrangements} = \frac{362880}{5040 \times 1 \times 1} = \frac{362880}{5040} = 72 \] ### Step 7: Calculate the Probability Now, we can find the probability \(P\) that the product of the digits equals 1920: \[ P = \frac{\text{Favorable cases}}{\text{Total cases}} = \frac{72}{5^9} \] ### Step 8: Final Result Calculating \(5^9\): \[ 5^9 = 1953125 \] Thus, the probability is: \[ P = \frac{72}{1953125} \] ### Conclusion The probability that the product of all digits in the nine-digit number equals 1920 is: \[ \frac{72}{1953125} \]
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