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I need help with the following problem: There are $3$ groups of people: Group $A$ of $4$ people, Group $B$ of $5$ people, Group $C$ of $6$ people. In how many ways can these people receive $5$ identical rewards if...

  1. At Least $2$ people from group $B$ and at least $2$ people from group $C$ must receive a prize.
  2. At least one person from each group must receive a reward, but no more then $2$ people from one group.

I know that the answer to 1. is $950$ and the answer to 2. is $1410$, but I would really like an explanation.

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If you enumerate these are the possibilities and I will do Part I and you complete Part II enter image description here

Part I Total Number of ways $$ = {4\choose1}{5\choose2}{6\choose2}+{4\choose0}{5\choose3}{6\choose2}+{4\choose0}{5\choose2}{6\choose3} = 950$$

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  • $\begingroup$ Thanks for Part 1! Is this correct for Part 2? imgur.com/QaljnDc $\endgroup$ – MDTech.us_MAN Nov 17 '17 at 8:04
  • $\begingroup$ You are right if you follow the same way of calculating $\endgroup$ – Satish Ramanathan Nov 17 '17 at 8:07
  • $\begingroup$ This seems like the most straight-forward way of doing it. Thanks for the help! $\endgroup$ – MDTech.us_MAN Nov 17 '17 at 8:11
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the answer to 1 is 1950 because: first we need to choose 2 people from B and 2 people from C so we have in total $5\choose{2} $$ 6\choose{2} $ and the we need to choose out of 13 people who are left means in total : 13*$5\choose{2} $$ 6\choose{2} $ = 1950

in answer 2 we get 1410 because: there are three cases all together, the first is that only one person from A gets a reward and 2 from set B and C similarly for each of those 3 sets then we get there are: $5\choose{2} $$ 6\choose{2} $*4+$4\choose{2} $$ 6\choose{2} $*5+$5\choose{2} $$ 4\choose{2} $*6 = 1410

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    $\begingroup$ Four people have already been chosen for receiving the gifts, how do you get 13. I don't think this method ensures that condition of B and C having atleast two gifts. I am trying to understand your method $\endgroup$ – Satish Ramanathan Nov 17 '17 at 8:33
  • $\begingroup$ It seems like if you were to select 3 people from group B, I say you can do it in ${5\choose3}$ and you say I select the 2 that is necessary in ${5\choose2}$ and the third person in ${3\choose1}$ ways to to total of ${5\choose3}{3\choose1}$ ways. Similarly for group C. In the first case you adding a factor of 3 and in the group C case you are adding a factor of 4. Will you let me know which you think is correct? $\endgroup$ – Satish Ramanathan Nov 17 '17 at 9:18
  • $\begingroup$ Your answer to the first question is wrong. First of all, once you have selected two people from group B and two group C, you have $11$ people left, not $13$. However, $11\binom{5}{2}\binom{6}{2}$ still over counts since you count distributions in which three people from the same group receive an award three times, once for each way you could designate two of the three recipients as the ones reserved for people in that group. $\endgroup$ – N. F. Taussig Nov 17 '17 at 10:17

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