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Lets suppose we have 16 different objects and we want to place them in four different boxes containing $4$ each . We need to find the probability of specific two objects lets say $a_1,a_2$ to be together in a box .

What i did was favourable cases would be:
$\binom{14}{2} * \binom{12}{4} * \binom{8}{4}*\binom{4}{4}$ * 4 , this is the case of those two to be lets say in first box and then other distribution in rest boxes , and similarily other three cases of those two to be in other three boxes . Sample space would be $\binom{16}{4} * \binom{12}{4} * \binom{8}{4}*\binom{4}{4}$ dividing both gives the required answer , but i would like if there is a quicker/shorter way to get the same answer .
This post has that method but i am not able to fully comprehend how it does : There are 15 balls named from A to O.

ProblemDestroyer
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  • I don't understand how one could get quicker than this. It is juts a multiplication problem once you have the formula – tryst with freedom May 15 '22 at 17:02
  • I mean alterante approach which doesnt lead counting like i did , as to why i am asking , see this problem where the solution given was way shorter than what my method or the op method from that post gives . https://math.stackexchange.com/q/3909366/1021792 @Aplateofmomos – ProblemDestroyer May 15 '22 at 17:05
  • Say the two specific objects are $A$ and $B$. Once you place object $A$ in a box, think of the placements as $A _ _ _ _ .._$. That is $15$ places for $B$ but only $3$ of them will lead to $B$ being in the same box as $A$. – Math Lover May 15 '22 at 17:10
  • 15 places for B?? something sounds wrong there @MathLover – tryst with freedom May 15 '22 at 17:12
  • $B$ can either be any of the $3$ objects in the same box as $A$ or it can be any of the $12$ objects in other three boxes. – Math Lover May 15 '22 at 17:17
  • @MathLover in my approach i considered the inside the boxes they are indistinguishable only which of them are what is considered ( that is order doesnt matter inside any box ) . Isnt ? – ProblemDestroyer May 15 '22 at 23:22
  • But your method is giving them places inside a box too , which i dont get properly – ProblemDestroyer May 15 '22 at 23:23

1 Answers1

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Look at my answer here which explains the method for a similar, but more complex problem solved by many methods.

The answer here is thus simply $\dfrac3{15} = \dfrac15$

PS

There they had to be placed in different groups, here they are to be placed in the same group, but the idea behind the solution is the same.

true blue anil
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  • Sir i understood , but how we will reasons that probability will come out to be same as the calcualtion usual one ? I mean without calculating values i mean to show there is a bijection in both methods to give sams answer of proabability – ProblemDestroyer May 19 '22 at 14:21
  • Whenever probability is asked for, it is often possible to avoid counting arrangements. Here you can define $E1=$ {event that $a_1$ is in some box}, and $E2=$ {event that $a_2$ is in the same box as $a_1$}. By the general multiplication rule, $P(E1E2) = P(E1)P(E2|E1) = 1*\frac{3}{15}$ – true blue anil May 19 '22 at 15:48
  • Understood thanks Sir :) – ProblemDestroyer May 19 '22 at 20:17