Question:
The number of seven-digit numbers that can be formed by using the digits \(1,2,3,5,7\) such that each digit is used at least once, is:
The number of seven-digit numbers that can be formed by using the digits \(1,2,3,5,7\) such that each digit is used at least once, is:
Updated On: Apr 10, 2026
- \(15400\)
- \(17800\)
- \(16800\)
- \(29400\)
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The Correct Option is C
Solution and Explanation
Concept:
We must form \(7\)-digit numbers using \(5\) digits with each digit appearing at least once.
This is solved using combinations and permutations with repetition.
Step 1:Distribute the extra digits.} Since each of the \(5\) digits must appear at least once, we first place one of each. Total digits used \(=5\) Remaining digits \(=7-5=2\) These \(2\) extra digits can be distributed among the \(5\) digits. Number of ways: \[ \binom{2+5-1}{5-1} = \binom{6}{4} = 15 \]
Step 2:Arrange the digits.} For each distribution, the total number of arrangements is \[ \frac{7!}{a_1!a_2!a_3!a_4!a_5!} \] where \(a_i\) are the frequencies of the digits. Summing across all possible distributions simplifies to \[ 16800 \] Thus the number of such seven-digit numbers is \[ \boxed{16800} \]
Step 1:Distribute the extra digits.} Since each of the \(5\) digits must appear at least once, we first place one of each. Total digits used \(=5\) Remaining digits \(=7-5=2\) These \(2\) extra digits can be distributed among the \(5\) digits. Number of ways: \[ \binom{2+5-1}{5-1} = \binom{6}{4} = 15 \]
Step 2:Arrange the digits.} For each distribution, the total number of arrangements is \[ \frac{7!}{a_1!a_2!a_3!a_4!a_5!} \] where \(a_i\) are the frequencies of the digits. Summing across all possible distributions simplifies to \[ 16800 \] Thus the number of such seven-digit numbers is \[ \boxed{16800} \]
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