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ENEM
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Luckisnoexcuse
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Edited the post. Agree with the solution below


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mynamegoeson
There will be 4C3 ways in which any three out of A B C D can be picked . there are 2C1 ways in which A can be picked . 4C1 ways in which B can be picked. 1C1 ways in which C can be picked and 1C1 ways in which D can be picked


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I am sorry but I do not agree with your explanation. If A is repeated twice, that doesn't mean number of ways of selecting 1 A is 2C1.

Since, the original question asks for a 3 letter code, which much be unique.

Case 1: When All letters are same: BBB. -- 1

Case 2: AA and one other letter. 1 * 3C1 * 3!/2! = 9 ways.

Case 3: BB and one other letter . 1 * 3C1 * 3!/2! = 9 ways.

Case 4: When all are different:

We can have any of the three selected from A,B,C and D = 4C3

Then we can arrange those 3 letters = 3!.

Total number of ways = 4C3 * 3! = 4 * 6 = 24.

Hence, answer is 1 + 9+9+24 = 43 ways.

Please let me know in case I missed anything.
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Hi,

thank you both abhimahna and mynamegoeson, for replying. I had dropped a mail regarding this to magoosh experts. 43 is the correct answer with correct explanation.
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