in Set Theory & Algebra edited by
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1 vote
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in Set Theory & Algebra edited by
618 views

4 Comments

@srestha ma'am divide by 2! Because in 6! We already counted {1,2} {3,4} as different permutation but both are not labelled so we have to remove them that's why we are dividing by 2! .
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Yes mam it's just same as "dividing into a groups "
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yes :)

@Prateek solved as balls and bin like problem

check :)
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1 Answer

3 votes
3 votes
Best answer

This question is same as solving 6 balls and unleveled container problem

Now, we can arrange those $6$ numbers in $4$ container as $2$ balls and $2$ balls in first two container and $1$ ball and $1$ ball in  last two container

Case $1:$

If we take $1$ st box contain $\left \{ 1,2 \right \}$ and $2$ nd box contain $\left \{ 3,4 \right \}$

$3$rd and $4$ th box contains $\left \{ 5 \right \},\left \{ 6 \right \}$

Then, total number of selection possible $\binom{6}{2}\times \binom{4}{2}\times \binom{2}{1}\times \binom{1}{1}$

but here $\left \{ 5 \right \},\left \{ 6 \right \}$ can arrange among themselves (here we need to divide by $2!$)

and $\left \{ 1,2 \right \}$ and  $\left \{ 3,4 \right \}$ can arrange among themselves, which are no need to be counted (here we need to divide by $2!$)

So, for $2,2,1,1$ total number of arrangement possible $\frac{\binom{6}{2}\times \binom{4}{2}\times \binom{2}{1}}{2!2!}=45$

Case $2:$

All box contain $2$ balls

And boxes are indistinguishable

So, total no. of arrangement possible $\frac{\binom{6}{2}\times \binom{4}{2}\times \binom{2}{2}}{3!}=15$

Note: Here we are doing arrangement of container , and not balls

So, from $(1)$ and $(2)$ we get $45+15=60$

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4 Comments

But I think partition problem can be solved in ball and bin problem

mam, i don't know what is  ball and bin problem

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@Shaik

go here https://classroom.gateoverflow.in/login/index.php

u get many links of balls and bin

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mam, that is only login page,

check the link one more time :)
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