Question

Seven digits from the numbers \(1,2,3,4,5,6,7,8\) and 9 are written in random order. The probability that this seven digit number is divisible by 9 is : (a) \(\frac{7}{9}\) (b) \(\frac{1}{9}\) (c) \(\frac{2}{9 !}\) (d) \(\frac{4}{9}\)

Short answer

Answer: The correct probability is \(\frac{1}{30240}\). None of the given options are correct.

Step-by-step solution

Find the total number of possible arrangements without the constraint

There are a total of 9 digits to choose from, and we want to choose 7 of them. The total number of ways to arrange these 7 digits without any constraint is given by the following formula: Total Arrangements = \(_9P_7 = \frac{9!}{(9-7)!}\) Calculate the total arrangements: Total Arrangements = \(\frac{9!}{2!} = \frac{9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3}{2} = 181440\)

Find the total number of ways to arrange the seven digits such that the sum of digits is divisible by 9

One digit must be excluded to create the seven-digit number. We can exclude one digit in 9 different ways, by removing a single digit from the set \(\{1,2,3,4,5,6,7,8,9\}\). By removing a digit, the sum of the remaining 7 digits should be divisible by 9. Sum of all nine digits: Sum = \(1+2+3+4+5+6+7+8+9 = 45\) Now, we need to find the number of ways to eliminate one digit such that the sum of the remaining seven digits is divisible by 9. There are three possibilities: 1. Remove a digit whose value is \(9k\) 2. Remove a digit whose value is \((9k-1)\) 3. Remove a digit whose value is \((9k-2)\) where k is an integer. Case 1: If we remove a digit whose value is \(9k\), the sum of the remaining digits should also be divisible by 9. We have one such number in the set: 9. Case 2: If we remove a digit whose value is \((9k-1)\), the sum of the remaining digits should be equal to \((9k-1)\). We have three such numbers in the set: 2, 5, and 8. Case 3: If we remove a digit whose value is \((9k-2)\), the sum of the remaining digits should be equal to \((9k-2)\). We have two such numbers in the set: 1 and 7. In total, there are \(1+3+2=6\) ways to remove a digit and make the sum of the remaining digits divisible by 9.

Calculate the probability

Now, we can calculate the probability by dividing the number of successful arrangements by the total possible arrangements: Probability = \(\frac{\text{Successful Arrangements}}{\text{Total Arrangements}} = \frac{6}{181440}\) Simplify and write the probability as a fraction: Probability = \(\frac{1}{30240}\) Finally, we compare our result with the given options: (a) \(\frac{7}{9}\) (b) \(\frac{1}{9}\) (c) \(\frac{2}{9 !}\) (d) \(\frac{4}{9}\) None of the given options match our result, which means that none of the options are correct. The correct probability is \(\frac{1}{30240}\).

Key concepts

Permutation

When talking about permutations, we're discussing different ways to arrange a set of objects. In this exercise, the term permutation refers to rearranging the selected digits to form distinct numbers.

• For example, if we have three digits: 1, 2, and 3, the permutations would be 123, 132, 213, 231, 312, and 321.
• Permutations consider the sequence of arrangement; thus, different orders count as different permutations.

In mathematical terms, to find the permutations of a certain number of items (n), we can use the formula:\[ _nP_r = \frac{n!}{(n-r)!} \]where \(!\) denotes a factorial, and in the current problem, we use this formula to determine the number of different 7-digit arrangements we can make by selecting from a larger set of 9 digits.

Divisibility Rules

Divisibility rules are a set of guidelines that help determine whether a number is divisible by another number without actually performing the division. These rules simplify the process, particularly when dealing with large numbers.
In our exercise, we are interested in the divisibility rule for 9:

• A number is divisible by 9 if the sum of its digits is divisible by 9.
• This means that how we arrange the numbers isn't what matters; the critical part is the sum of those numbers.

In the problem, after removing one digit from the entire set, we need the remaining digit's sum to be divisible by 9 for the resultant number to be divisible by 9. By understanding and applying these simple rules, we can quickly identify whether the number meets the criteria.

Combinations

Combinations pertain to selecting items from a larger set where the order of selection doesn't matter. This is slightly different from permutations where order does matter.

• For example, choosing two fruits from a group of an apple, banana, and cherry is the same whether you pick the apple and then the banana or the banana and then the apple.
• The formula for combinations is:\[ _nC_r = \frac{n!}{r!(n-r)!} \]

In the context of our problem, we're picking seven digits from a total of nine available digits to form a number. However, to ensure it's divisible by 9, we consider the sum using divisibility rules, going beyond mere combinations.

Factorial

The concept of factorial is integral to permutations and combinations. A factorial is the product of an integer and all the integers below it, down to one.

• For instance, the factorial of 5 (written as \(5!\)) is calculated as: 5 x 4 x 3 x 2 x 1, which equals 120.
• In mathematical notation, \(n!\) implies multiplying all whole numbers from the number "n" down to 1.

The factorial function is crucial when calculating permutations and combinations because it helps determine the number of possible ways to arrange or select a set of items. In this exercise, factorial calculations allow us to compute the total number of arrangements efficiently.