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22 Jun 2023, 07:42
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Given: A 3-digit positive integer consists of non-zero digits and 2 of the digits are the same.
Asked: How many such integers are possible?
Solution: There can be a total of only 3 cases -> ABA or AAB or BAA
Now we have 1-9 ( 9 digits as 0 not included) to fill.
So the Total Cases would be -> 9 * 8 ( As A can be filled in 9 ways and B in 8 ways ) * 3 ( as there are a total of 3 cases ) = 216.
Answer: 216 (C).
Asked: How many such integers are possible?
Solution: There can be a total of only 3 cases -> ABA or AAB or BAA
Now we have 1-9 ( 9 digits as 0 not included) to fill.
So the Total Cases would be -> 9 * 8 ( As A can be filled in 9 ways and B in 8 ways ) * 3 ( as there are a total of 3 cases ) = 216.
Answer: 216 (C).
12 Dec 2024, 09:01
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Hello Mam!
Thanks for the solution. But, I have one confusion in this. For X we have 9 ways, for Y we have 8 ways. But third digit can be either X or Y so, we have to multiply 9*8*2*3 for the correct answer.
Please guide me where am I getting confused in this.
Thanks for the solution. But, I have one confusion in this. For X we have 9 ways, for Y we have 8 ways. But third digit can be either X or Y so, we have to multiply 9*8*2*3 for the correct answer.
Please guide me where am I getting confused in this.
KarishmaB
16 Dec 2025, 06:46
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