1 votes 1 votes 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. Combinatory combinatory counting + – Pineapple asked Aug 25, 2022 • retagged Aug 25, 2022 by makhdoom ghaya Pineapple 591 views answer comment Share Follow See all 7 Comments See all 7 7 Comments reply Kabir5454 commented Aug 25, 2022 reply Follow Share i am getting 27 . 0 votes 0 votes Pineapple commented Aug 25, 2022 reply Follow Share How are you solving it? Can you describe your approach? I’m not even getting how to approach this problem. 0 votes 0 votes Kabir5454 commented Aug 25, 2022 reply Follow Share is it 27 can you confirm the answer ? 0 votes 0 votes shishir__roy commented Aug 25, 2022 i edited by shishir__roy Aug 25, 2022 reply Follow Share @Kabir5454 I'm also getting $27$.Let $c_n$ be the number of candies recieved by $n^{th}$ child.$c_1 + c_2 + c_3 = 10$And conditions are $c_1 \ge 2, c_2 \le 6 \text{ and } c_3 \le 3$.So, we'll first give 2 choclates to $c_1$.Now our equation becomes $c_1 + c_2 + c_3 = 8$And conditions are $c_2 \le 6 \text{ and } c_3 \le 3 \text{ and } c_1 \ge 0$ (trivial condition).Now, using complement rule and inclusion exclusion principle -$ans = total - (c_2 \ge 7 \text{ or } c_3 \ge 4)$$ans = total - ((c_2 \ge 7) + (c_4 \ge 4) - (c_2 \ge 7 \text{ and } c_3 \ge 4))$$ans = {10 \choose 8} - ({3 \choose 1} + {6 \choose 4} - 0) = 45 - (3+15) = 27$. 4 votes 4 votes Kabir5454 commented Aug 25, 2022 reply Follow Share Yes i have another approach as well . 0 votes 0 votes shishir__roy commented Aug 25, 2022 reply Follow Share @Kabir5454 Nice, your answers are always enlightening 1 votes 1 votes Pineapple commented Aug 25, 2022 reply Follow Share yes it is correct 0 votes 0 votes Please log in or register to add a comment.
Best answer 8 votes 8 votes 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 Kabir5454 answered Aug 25, 2022 • edited Aug 25, 2022 by Kabir5454 Kabir5454 comment Share Follow See all 0 reply Please log in or register to add a comment.
1 votes 1 votes im getting answer as 27 [ Jiren ] answered Aug 25, 2022 [ Jiren ] comment Share Follow See all 3 Comments See all 3 3 Comments reply shishir__roy commented Aug 25, 2022 reply Follow Share You over complicated the solution. 0 votes 0 votes [ Jiren ] commented Aug 25, 2022 reply Follow Share @shishir__roy just completed till lecture 10 bro 😅 so i dont know other ways to solve than using IEP 0 votes 0 votes shishir__roy commented Aug 25, 2022 reply Follow Share Read my comment above. 0 votes 0 votes Please log in or register to add a comment.