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Bunuel
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Combinations of 3-digit codes: All three digits (1, 2 and 3) must be used to get a total of 6.
Hence, 3! = 6
Combinations of 2-digit codes: Only 3+3 works. Hence, 1.

Total = 6 + 1 = 7

VibhuAnurag 2-2-2 is also a way to get a total of six, so I think the correct answer would be 8.

Yes Benji, you are right. Considering 2+2+2 = 6, it will be 8. Corrected.
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Bunuel
A password generator creates 2- or 3-digit codes. How many different codes can it generate using only the digits 1, 2, or 3, if the sum of each code's digits must be 6?

A. 7
B. 8
C. 9
D. 10
E. 11



Oa is B

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Bunuel
A password generator creates 2- or 3-digit codes. How many different codes can it generate using only the digits 1, 2, or 3, if the sum of each code's digits must be 6?

A. 7
B. 8
C. 9
D. 10
E. 11



To get a sum of 6, you will need to enumerate though it is easy to do so.

Two digit codes __ __
To get a sum of 6, both digits must be 3 only. There is only 1 such code.

Three digit codes __ __ __
1+2+3 = 6 so using the three digits exactly once will give us a sum of 6. We can get 3! codes in this case.
Can we make 6 with another combination in which we use 3 once and some 1s and 2s? No. For a sum of 6, when we have one 3, the other two digits should also add up to 3 so they must be 1 + 2.

We cannot use 3 twice or more because then the sum will exceed 6 with the third digit.

Can we use 3 zero times i.e. can we get 6 using only 1s and 2s? The sum in this case will vary from 3 (all 1s) to 6 (all 2s). Since we need the sum to be exact 6, another code is 2+2+2. Nothing else is possible.

Total = 1 + 3! + 1 = 8

Answer (B)

Check out this video on Permutations: https://youtu.be/LFnLKx06EMU
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Approach: Split into cases based on number of digits, then find all valid digit combinations.

Case 1: 2-digit codes
We need two digits from {1, 2, 3} that sum to 6.

The only possibility: 3 + 3 = 6
Code: 33

Count for 2-digit codes: 1

Case 2: 3-digit codes
We need three digits from {1, 2, 3} that sum to 6.

Option A: 1 + 2 + 3 = 6
Since all three digits are different, we can arrange them in 3! = 6 ways:
123, 132, 213, 231, 312, 321

Option B: 2 + 2 + 2 = 6
All digits are the same, so only 1 arrangement:
222

Count for 3-digit codes: 6 + 1 = 7

Total codes = 1 + 7 = 8

Answer: B
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