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Find the number of positive integers that satisfy both the following conditions:

Each digit is a 1 or a 2

The sum of the digits is 10

 Feb 12, 2023
 #1
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If we have a number with all 1s, the maximum number of digits we can have is 5, as 1 + 1 + 1 + 1 + 1 = 5. If we have a number with all 2s, the maximum number of digits we can have is 5, as 2 + 2 + 2 + 2 + 2 = 10. If we have a mix of 1s and 2s, the maximum number of digits we can have is 4, as 1 + 2 + 2 + 2 = 7.

 

Let's list out the possibilities:

 

Case 1: All 1s The number of possibilities here is the number of ways to choose 5 digits out of 5 digits, which is C(5, 5) = 1.

 

Case 2: All 2s The number of possibilities here is the number of ways to choose 5 digits out of 5 digits, which is C(5, 5) = 1.

 

Case 3: Mix of 1s and 2s The number of possibilities here is the sum of the number of ways to choose i digits out of 4 digits, where i ranges from 1 to 4, inclusive. That is, C(4, 1) + C(4, 2) + C(4, 3) + C(4, 4) = 4 + 6 + 4 + 1 = 15

 

Adding up the number of possibilities in each case, we have 1 + 1 + 15 = 17. Therefore, there are 17 positive integers that satisfy both conditions.

 Feb 12, 2023
 #2
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hmm, it seems that the answer was incorrect, but your reasoning seems fine to me. I'm not sure what went wrong.

Guest Feb 12, 2023

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