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How many ways are there to distribute 10 identical candies among 3 children such that the first child receives at least 2 candies, the second child receives atmost 6 candies and the third child receives atmost 3 candies.
in Combinatory retagged by
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4 Comments

Yes i have another approach as well .
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@Kabir5454 Nice, your answers are always enlightening

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yes it is correct
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2 Answers

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7 votes
Best answer

Lets frame this problem using generating function .

The given Problem is

$x1+x2+x3=10$ where , $x1 \geq 2$ ,$x2 \leq 6$ ; $x3 \leq3$ ;

Using generating function this can be written as ,

$(x^{2}+x^{3}+x^{4}+x^{5}+x^{6}+x^{7}+x^{8}+x^{9}+x^{10})(1+x+x^{2}+x^{3}+x^{4}+x^{5}+x^{6})(1+x+x^{2}+x^{3})$

How did i wriite it ?

=> write the possible value $x1$ can take in power of $x$ . same do for $x2,x3$ .

Now our problem reduces to only we need to find the coefficient of the $x^{10}$ which is the required answer.

Simplifying the expression :-

=$\large \frac{x^{2}(1-x^{9})}{1-x}*\frac{1-x^{7}}{1-x}*\frac{1-x^{4}}{1-x}$

=$\large (x^{2}-x^{11})(1-x^{7})(1-x^{4})(1-x)^{-3}$

as we need only power of $x^{10}$ so while simplifying we ignore power of the term which is greater that 10.

=$\large (x^{2}-x^{6}-x^{9})(1-x)^{-3}$

we from $\large (1-x)^{-3}$ we need only coefficient of $\large x,x^{4},x^{8}$.

$\large (1-x)^{-3}$=$(1+3x+..+15x^{4}+..+45x^{8}…….)$

So coefficient  of $x^{10}$= $(45-15-3)=27$

 

This also an another efficient approach :- https://gateoverflow.in/381380/combinatorics?show=381390#c381390

edited by
1 vote
1 vote

im getting answer as 27 

 

3 Comments

You over complicated the solution.
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@shishir__roy just completed till lecture 10 bro 😅

so i dont know other ways to solve than using IEP

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Read my comment above.
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