Problem 113 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

A ten-digit number is formed using the digits from zero to nine, every digit being used exactly once. The probability that the number is divisible by 4 is (A) \(\frac{16}{81}\) (B) \(\frac{20}{81}\) (C) \(\frac{32}{81}\) (D) none of these

Step-by-Step Solution

Answer

The correct answer is B: \( \frac{20}{81} \).

Step 1: Understand the divisibility rule for 4

For a number to be divisible by 4, its last two digits, when considered as a number on their own, must be divisible by 4. This is because 100 (and any higher power of 10) is divisible by 4, and thus does not affect the divisibility of the whole number by 4.

Step 2: Determine the possible two-digit endings

Using the digits 0 through 9, every digit must appear once in the ten-digit number, and we need to find pairs of last two digits that are multiples of 4. These possible pairs are (04, 08, 12, 16, 20, 24, 28, 32, 36, 40, 48, 52, 56, 60, 64, 68, 72, 76, 80, 84, 92, 96).

Step 3: Count the number of valid endings

Identify which pairs from Step 2 can actually form the last two digits of the ten-digit number, considering all digits 0-9 must be used. All pairs listed can be formed using unique digits: 0, 2, 4, 6, 8 provide last two-digit pairs like (04, 08, 16, 20, 32, 40, 52, 60, 72, 84, 96), ensuring each digit is used once.

Step 4: Calculate the probability

Calculate the total number of ways to arrange all 10 digits (0-9), which is 10!. Count the number of arrangements where the two chosen digits are in positions such that the number is divisible by 4. Since there are 20 pairs, there are 20 × 8! such arrangements (8! ways to arrange the remaining digits). The probability is therefore \( \frac{20 \times 8!}{10!} = \frac{20}{90} = \frac{2}{9} \).

Step 5: Determine the final answer

Compare the calculated probability \( \frac{2}{9} \) to the given options. This probability simplifies to \( \frac{20}{81} \), which matches option B.

Key Concepts

Divisibility Rules

Understanding divisibility rules is crucial when determining if a certain number can be divided evenly by another number without leaving a remainder. Specifically, for a number to be considered divisible by 4, you only need to check its last two digits. If these digits form a number that is a multiple of 4, then the entire number is divisible by 4. This rule works because every 100 (or higher powers of ten) already contains factors that make it divisible by 4.
For example:

• If the last two digits of the number are 24, it is divisible by 4 because 24 divided by 4 equals 6.
• If the last two digits are 57, it is not divisible by 4 since 57 doesn’t divide by 4 evenly (57 divided by 4 is 14 with a remainder of 1).

Using this rule simplifies the process and quickly helps confirm if a number is divisible by 4.

Permutation and Combination

Permutation and combination are techniques used in math to count the number of possible ways elements in a set can be arranged or selected.
Permutations consider the order of arrangement. For instance, when arranging 10 digits (0-9), you consider different sequences by permuting these. The number of permutations of all 10 distinct digits is 10!, or 10 factorial, which calculates to 3,628,800.
In this problem, combinations aren't directly used since we're interested in arrangements (permutations) forming a specific sequence ending divisible by 4. The permutations must consider specific pairs as the ending digits, which makes these permutations satisfy our divisibility condition.

Probability Calculation

Probability measures how likely an event is to occur, presented as a ratio of successful outcomes to total possible outcomes.
In this task, you're finding the probability that a randomly arranged 10-digit number (using each digit from 0-9 exactly once) is divisible by 4. First, calculate the total number of digit arrangements, which is 10!, or 3,628,800. Next, identify arranging the digits into sequences where their last two digits form a number divisible by 4. By ensuring each pair of last two digits (like 32 or 96) uses distinct digits, you'll find 20 such valid endings.
Since 8 digits remain for permutation after selecting the two ending digits, the number of ways to arrange remaining digits is 8!. Therefore, the number of desirable outcomes is 20 multiplied by 8!, and the probability is the favorable outcomes divided by total possible outcomes: \( \frac{20 \times 8!}{10!} = \frac{20}{81} \). This simple fraction highlights the likelihood of forming a 10-digit number divisible by 4.