Letters in boxes - combinations

Question

Q .) The number of ways can 5 letter be put in 3 boxes A, B,C such that A has at least 2 letters.

 

My approach: Number of ways to choose  2 letters out of 5 is 5C2. And for each such combination the remaining 3 letter have 3 choice.

Therefore 5C2 * 3^3, but this is incorrect. Please point out the fault in my understanding and also the correct way to solve it

Comments

But why left with only 2 boxes? They have asked for A to be containing at least 2, not exactly 2.

Please explain what am I misinterpreting?

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Ok I got my mistake. Thanks @balchandar reddy san bhai.

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Here letters are diff and also boxes are diff, So, arrangement of letter in box also should be counted.right??

Answer 1

A               B   +   C

0                 5                (2^5)   => 5C0* (2^5) = 32

1                 4                (2^4)   => 5C1* (2^4) = 80

2                 3                (2^3)   => 5C2* (2^3) = 80

3                 2                (2^2)   => 5C3* (2^2) = 40

4                 1                (2^1)   => 5C4* (2^1) = 10

5                 0                (2^0)   => 5C5* (2^0) = 1

Total                            243

With A>=2, 80 + 40 + 10 + 1 = 131

Explanation:

Let me explain the case where A = 2 and B + C = 3

we can chose 2 out of 5 letters in 5C2 ways and now B+C will have 3 letters which can be put in 2^3 ways

Comments

So in general we could do 3^5 - ( 5C0* (2^5) + 5C1* (2^4)).

Thanks sir :)

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@balchandar reddy san

Is repetition allowed here??