All Exams Test series for 1 year @ ₹349 only
Question

Consider the digits 3, 5, 7, 9. What is the number of 5-digit numbers formed by these digits in which each of these four digits appears?

The correct answer is

240

Forming 5-Digit Numbers Using Digits 3, 5, 7, 9

The problem asks us to find the number of 5-digit numbers that can be formed using the digits 3, 5, 7, and 9, with the condition that each of these four digits must appear at least once in the 5-digit number.

We are forming a 5-digit number using a set of 4 distinct digits {3, 5, 7, 9}. Since the number has 5 digits and we must use all 4 distinct digits, exactly one of the digits must be repeated.

Let the five positions for the digits be P1, P2, P3, P4, P5. We need to fill these positions using the digits 3, 5, 7, 9 such that all four digits are present. This means the set of 5 digits used will contain the four distinct digits {3, 5, 7, 9} plus one extra digit, which must be a repeat of one of the four.

There are 4 possibilities for which digit is repeated:

  • Digit 3 is repeated. The digits used are {3, 3, 5, 7, 9}.
  • Digit 5 is repeated. The digits used are {3, 5, 5, 7, 9}.
  • Digit 7 is repeated. The digits used are {3, 5, 7, 7, 9}.
  • Digit 9 is repeated. The digits used are {3, 5, 7, 9, 9}.

Now, for each case, we need to find the number of distinct permutations of the 5 digits. The formula for permutations of $n$ objects where one object appears $r$ times is $\frac{n!}{r!}$. In our cases, $n=5$ (total digits) and $r=2$ (the repeated digit appears twice).

Let's calculate the number of arrangements for each case:

Repeated Digit Set of Digits Number of Permutations
3 {3, 3, 5, 7, 9} $\frac{5!}{2!} = \frac{120}{2} = 60$
5 {3, 5, 5, 7, 9} $\frac{5!}{2!} = \frac{120}{2} = 60$
7 {3, 5, 7, 7, 9} $\frac{5!}{2!} = \frac{120}{2} = 60$
9 {3, 5, 7, 9, 9} $\frac{5!}{2!} = \frac{120}{2} = 60$

Since these four cases are mutually exclusive (a number cannot have both 3 and 5 as the only repeated digit), the total number of 5-digit numbers formed by these digits where each of the four digits appears is the sum of the numbers of permutations in each case.

Total number of 5-digit numbers = (Permutations when 3 is repeated) + (Permutations when 5 is repeated) + (Permutations when 7 is repeated) + (Permutations when 9 is repeated)

Total number = $60 + 60 + 60 + 60 = 4 \times 60 = 240$.

Thus, there are 240 such 5-digit numbers.

Revision Table: Key Concepts

Concept Description Application in this problem
Permutations Arrangements of objects where order matters. Used to arrange the 5 digits (with one repeated).
Permutations with Repetition Formula $\frac{n!}{r!}$ for arranging $n$ items where one item repeats $r$ times. Used to calculate arrangements for each set of 5 digits.
Mutually Exclusive Cases Cases that cannot happen at the same time. The cases based on which digit is repeated are mutually exclusive.

Additional Information: Combinations vs. Permutations

It's important to understand the difference between combinations and permutations when solving problems like this.

  • Combinations: Refer to the selection of objects where the order does not matter. For example, choosing a committee of 3 people from a group of 10.
  • Permutations: Refer to the arrangements of objects where the order does matter. For example, arranging 3 people in a line or forming a number using digits.

In this problem, we are forming 5-digit numbers, and the order of the digits matters (e.g., 33579 is different from 97533). Therefore, we use permutations.

The condition that each of the four digits (3, 5, 7, 9) must appear means we cannot form numbers using only three or fewer distinct digits from the given set. Since we need a 5-digit number, and we start with 4 distinct digits, one digit must necessarily be used twice. Identifying this repeated digit is the first step in setting up the cases for permutations.

Was this answer helpful?

Important Questions from Fundamental Principles of Counting

  1. What is the number of four digit decimal number (<1) in which no digit is repeated?

  2. Let S = {2, 3, 4, 5, 6, 7, 9}. How many different 3-digit numbers (with all digits different) from S can be made which are less than 500?

  3. 3-digit numbers are formed using the digits 1, 3, 7 without repetition of digits. A number is randomly selected. What is the probability that the number is divisible by 3?

  4. Consider the following paragraph:

    THE ABILITY TO REASON ACCURATELY IS VERY IMPORTANT, AS IS THE ABILITY TO COUNT. AS AN EXERCISE IN BOTH, LET US COUNT HOW MANY TIMES THE LETTER "E" OCCURS IN THIS PARAGRAPH. THE CORRECT COUNT IS ________.

    Which option when put in the blank in the above paragraph will make the final sentence accurate?

  5. In an examination containing 10 questions, each correct answer is awarded 2 marks, each incorrect answer is awarded −1 and each unattampted question is awarded zero. Which of the following CANNOT be a possible score in the examination?

Need Expert Advice?

Start Your Preparation with Prepp Mobile App

Download the app from Google Play & App Store
Download the app from Google Play & App Store
Prepp Mobile App