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The coefficient of $x^{12}$ in $\left(x^{3}+x^{4}+x^{5}+x^{6}+\dots \right)^{3}$ is ___________.
in Combinatory retagged by
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4 Comments

Very good method @Sourav.Sharma

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The problem can read as :

You are given 3 buckets and you need to pick out 12 items, but each bucket only allows you to pick at least 3 items in one go (not less than that)

We can easily shift the origin from 3 to 0 (as 3 items are always compulsory to pick). So 9 will be picked anyway. We can only decide on the remaining 3.

And now we only need to pick 3 items (and we are allowed to pick any number of items starting from 1, 2,)
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17 Answers

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(x3 + x4 +x5 +x6 + ………...) (x3 + x4 +x5 +x6 + ………...) (x3 + x4 +x5 +x6 + ………...)

after multiplying we get the equation as = ax9 + bx10 +cx11 +dx12 + ex15+.……………….

 

coefficient of x9 =1 (only one possibe way 3+3+3)

coefficient of x10 =3 ( possibe ways 3+3+4,

                                                          3+4+3

                                                            4+3+3)

 

for coefficient of x12

just add the possible powers  to get 12.

3+3+6

3+4+5

3+5+4

3+6+3

 

 

4+3+5

4+4+4

4+5+3

 

5+3+4

5+4+3

 

6+3+3

 

there are 10 possible ways to get x12 term. i.e., coeffiecient is 10
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