Q.6. The number of seven-digit numbers, that can be formed by using the digits 1, 2, 3, 5 and 7 such

Q: What is the correct answer for JEE Main Mathematics Q6?

The correct answer is C. Sol. We have 5 distinct digits: 1, 2, 3, 5, 7. We need to form a 7-digit number using these digits, with each digit used at least once. This means two digits must be repeated. Case 1: Exactly one digit is repeated three times, and the other four digits are used once. Choose the digit to be repeated three times: $^5C_1$ ways. Arrange the 7 digits (3 identical, 4 distinct): $\frac{7!}{3!}$ ways. Num

Q: Q.6. The number of seven-digit numbers, that can be formed by using the digits 1, 2, 3, 5 and 7 such that each digit is used at least once, is :

The correct answer is C. Sol. We have 5 distinct digits: 1, 2, 3, 5, 7. We need to form a 7-digit number using these digits, with each digit used at least once. This means two digits must be repeated. Case 1: Exactly one digit is repeated three times, and the other four digits are used once. Choose the digit to be repeated three times: $^5C_1$ ways. Arrange the 7 digits (3 identical, 4 distinct): $\frac{7!}{3!}$ ways. Num

Q: What are the options for this JEE Main Mathematics question?

A. 15400 · B. 17800 · C. 16800 · D. 29400

Question

Q.6. The number of seven-digit numbers, that can be formed by using the digits 1, 2, 3, 5 and 7 such that each digit is used at least once, is :

FirstInTest · Accepted Answer

Sol. We have 5 distinct digits: 1, 2, 3, 5, 7. We need to form a 7-digit number using these digits, with each digit used at least once.
This means two digits must be repeated.
Case 1: Exactly one digit is repeated three times, and the other four digits are used once.
Choose the digit to be repeated three times: $^5C_1$ ways.
Arrange the 7 digits (3 identical, 4 distinct): $\frac{7!}{3!}$ ways.
Number of ways = $^5C_1 	imes \frac{7!}{3!} = 5 	imes \frac{5040}{6} = 5 	imes 840 = 4200$.

Case 2: Exactly two digits are repeated twice each, and the other three digits are used once.
Choose the two digits to be repeated twice: $^5C_2$ ways.
Arrange the 7 digits (2 identical, 2 identical, 3 distinct): $\frac{7!}{2!2!}$ ways.
Number of ways = $^5C_2 	imes \frac{7!}{2!2!} = 10 	imes \frac{5040}{4} = 10 	imes 1260 = 12600$.

Total number of such seven-digit numbers = $4200 + 12600 = 16800$.

Answer

15400

Answer

17800

Answer

16800

Answer

29400

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