Question:
How many numbers between \(10\) and \(10,000\) can be formed by using the digits \(1,2,3,4,5\), if no digit is repeated in any number?
How many numbers between \(10\) and \(10,000\) can be formed by using the digits \(1,2,3,4,5\), if no digit is repeated in any number?
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When numbers are to be formed within a range, first identify the possible number of digits, then count each case separately using permutations.
Updated On: Jun 26, 2026
- \(200\)
- \(775\)
- \(60\)
- \(120\)
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The Correct Option is A
Solution and Explanation
Step 1: Understand the range of numbers.
The numbers must lie between \[ 10 \] and \[ 10000 \] Therefore, the required numbers may be \(2\)-digit, \(3\)-digit, or \(4\)-digit numbers.
Step 2: Identify the available digits.
The available digits are \[ 1,2,3,4,5 \] No digit is repeated in any number.
Since there is no \(0\), there is no restriction on the first digit.
Step 3: Count the \(2\)-digit numbers.
For a \(2\)-digit number, choose and arrange \(2\) digits from \(5\) digits.
So, the number of \(2\)-digit numbers is \[ {}^5P_2 \] \[ {}^5P_2=5\times 4=20 \]
Step 4: Count the \(3\)-digit numbers.
For a \(3\)-digit number, choose and arrange \(3\) digits from \(5\) digits.
So, the number of \(3\)-digit numbers is \[ {}^5P_3 \] \[ {}^5P_3=5\times 4\times 3=60 \]
Step 5: Count the \(4\)-digit numbers.
For a \(4\)-digit number, choose and arrange \(4\) digits from \(5\) digits.
So, the number of \(4\)-digit numbers is \[ {}^5P_4 \] \[ {}^5P_4=5\times 4\times 3\times 2=120 \]
Step 6: Add all possible cases.
Total number of required numbers is \[ {}^5P_2+{}^5P_3+{}^5P_4 \] \[ =20+60+120 \] \[ =200 \]
Step 7: Final conclusion.
Therefore, \[ \boxed{200} \]
The numbers must lie between \[ 10 \] and \[ 10000 \] Therefore, the required numbers may be \(2\)-digit, \(3\)-digit, or \(4\)-digit numbers.
Step 2: Identify the available digits.
The available digits are \[ 1,2,3,4,5 \] No digit is repeated in any number.
Since there is no \(0\), there is no restriction on the first digit.
Step 3: Count the \(2\)-digit numbers.
For a \(2\)-digit number, choose and arrange \(2\) digits from \(5\) digits.
So, the number of \(2\)-digit numbers is \[ {}^5P_2 \] \[ {}^5P_2=5\times 4=20 \]
Step 4: Count the \(3\)-digit numbers.
For a \(3\)-digit number, choose and arrange \(3\) digits from \(5\) digits.
So, the number of \(3\)-digit numbers is \[ {}^5P_3 \] \[ {}^5P_3=5\times 4\times 3=60 \]
Step 5: Count the \(4\)-digit numbers.
For a \(4\)-digit number, choose and arrange \(4\) digits from \(5\) digits.
So, the number of \(4\)-digit numbers is \[ {}^5P_4 \] \[ {}^5P_4=5\times 4\times 3\times 2=120 \]
Step 6: Add all possible cases.
Total number of required numbers is \[ {}^5P_2+{}^5P_3+{}^5P_4 \] \[ =20+60+120 \] \[ =200 \]
Step 7: Final conclusion.
Therefore, \[ \boxed{200} \]
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