How many 3-digit numbers are possible using permutations without repetition of digits if digits are 1-9?
a) 504
b) 729
c) 1000
d) 720
The correct answer and explanation is :
Correct Answer: d) 720
Explanation:
We are asked to determine how many 3-digit numbers can be formed without repeating digits, using the digits from 1 to 9.
Step 1: Understand the Range of Digits
The digits available are 1 through 9, inclusive. That gives us:
$$
\text{Total digits available} = 9 \text{ digits} \quad (\text{Digits: } 1, 2, 3, 4, 5, 6, 7, 8, 9)
$$
Note: 0 is not included, so we don’t have to worry about a leading zero or its inclusion at all.
Step 2: Understand the Requirement
We are forming 3-digit numbers using permutations without repetition. That means:
- The order matters (since 123 is different from 321).
- No digit repeats in any number.
So, we are choosing 3 digits from the 9 available and arranging them in all possible ways.
Step 3: Use the Permutation Formula
The number of permutations without repetition of r items from n total items is given by:
$$
P(n, r) = \frac{n!}{(n – r)!}
$$
Here:
- $n = 9$ (digits from 1 to 9),
- $r = 3$ (we want 3-digit numbers).
So,
$$
P(9, 3) = \frac{9!}{(9 – 3)!} = \frac{9!}{6!}
$$
Now, calculate:
$$
P(9, 3) = \frac{9 \times 8 \times 7 \times 6!}{6!} = 9 \times 8 \times 7 = 504
$$
Wait — that gives 504, not 720. What’s wrong?
Let’s double-check. We’re being asked how many 3-digit numbers can be formed using digits 1–9 without repeating digits.
Let’s try to count them step-by-step instead.
Step 4: Count the 3-digit numbers manually
- For the hundreds place, we can use any of the 9 digits (1 through 9) – since 0 is not in the digit set.
- For the tens place, we can use 8 remaining digits (excluding the one already used).
- For the units place, we can use 7 remaining digits.
So:
$$
9 \times 8 \times 7 = 504
$$
So the correct answer is actually: a) 504, not d) 720.
✅ Corrected Answer: a) 504
Final Note:
The answer 720 would have been correct if the digits 0–9 were allowed and repetition was not allowed. But since the digits are only 1–9, the number of permutations of 3 digits chosen from 9 (without repetition) is 504.