The Number of Six-Digit Numbers That Can Be Formed Using the Digits 1, 3, 5, 7, 9 Such That Each Digit Is Used at Least Once Is

JEE Main 2020 — 7 January, Morning Shift. Previous Year Question.

Problem

The six-digit numbers that can be formed using digits $1,3,5,7,9$ such that each digit is used at least once — how many such numbers are there?

Key insight. With only $5$ distinct digits available for $6$ positions, “each digit used at least once” forces exactly one digit to be used twice and no digit to be skipped — there’s no other way to fill six slots with five required digits. That single observation turns an open-ended counting problem into a simple two-step choice.

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Approach

Since there are $5$ distinct digits and $6$ positions, exactly one digit must be repeated exactly once (used twice), while the other four digits appear exactly once. Choose which digit is the repeated one, then count the arrangements of the resulting six symbols (with one digit repeated twice) using the standard formula for permutations with repetition.

Solution

Step 1 — Determine the repetition structure

With $5$ digits and $6$ positions, and every digit required at least once, exactly one digit must appear twice, and the remaining four digits appear once each (no digit can appear three or more times, since that would leave fewer than $4$ slots for the other $4$ required digits).

Step 2 — Choose which digit repeats

$$\binom{5}{1} = 5 \text{ ways}$$

Step 3 — Arrange the six symbols (with one digit repeated twice)

For any fixed choice of the repeated digit, the number of distinct arrangements of $6$ symbols where one specific symbol appears twice is:

$$\frac{6!}{2!} = \frac{720}{2} = 360$$

Step 4 — Multiply

$$5\times360 = 1800$$

Answer

Common mistakes

• Forgetting to divide by $2!$ when arranging the six symbols. Since two of the six positions hold the same digit, swapping those two positions doesn’t create a new number — treating all six symbols as distinguishable and using plain $6!$ overcounts every valid arrangement exactly twice.
• Allowing a digit to repeat more than twice. With only $6$ positions and $5$ digits each required at least once, there’s no room for any digit to appear three or more times — trying to account for such cases (e.g. one digit appearing three times, another zero times) would violate the “each digit at least once” condition and shouldn’t be counted at all.

Practise next

• How many seven-digit numbers can be formed using the digits $1,2,3,4,5,6$ such that each digit is used at least once? Use the same choose-the-repeat-then-arrange approach.