Counting

Question

How many ways are there to put six temporary employees into four labeled offices so that there is at least one temporary employee in each of these four offices?

Answer 1

1 temporary employee in each office

So after giving each office 1 employee, there are remaining 2 employees.

Now we have to distribute 2 employee in 4 offices

$O_{1}$         $O_{2}$         $O_{3}$          $O_{4}$

   2                      0                   0                 0

   0                      2                    0                0

   0                      0                    2                0

   0                      0                    0                2

   1                      1                     0                0

   1                      0                     1                 0

   1                      0                      0                1

    0                     1                      1                 0

    0                      1                      0                 1

    0                       0                     1                  1

 

So, in 10 ways

Comments

@srestha,each employee and Office is distinguishable from each other.what you have derived is the case when both are unlabelled.

I am getting total 1020 ways.

(3-1-1-1) arrangements---->480 ways

(2-2-1-1) arrangments------>540 ways

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@reena

it will better if u explain, how u got 1020 , in more clear way

No, I have taken offices distinguishable , but employees undistinguishable,

Because here we cannot say emp1 is short highted than emp2 or something like that

But we tell there are 2 employees in office1, 2 employees in office2 ............like that

So, position of employee doesnot matter here. Only number of employee in an office matters

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@srestha, I am taking each empoyee different from each other.there are two possible arrangements then.

1)any 3 employess in one office and 1 in each remaining office(3-1-1-1)--->(choose 3  out of 6)*4!/3!=80 ways

2)for(2-2-1-1) arrangements---->⁶C₂*⁴C₂*(4!/2!2!)=540

total ways==80+540=620 ways.

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@reena_kandari

i am getting 480+1080 = 1560

$_{3}^{6}\textrm{C}$*4*3!  +  $_{2}^{4}\textrm{C}*_{2}^{6}\textrm{C}*_{2}^{4}\textrm{C}*2!$ = 1560

Answer 2

Let employees in office i = Oi

Therefore : O1 + O2 +O3+ O4 = 6  where Oi>=1

Putting at least 1  employee in each office we are left with (Oi complement = oi)

o1 + o2 +o3+ o4 = 6 - 4  where oi>=0

o1 + o2 +o3+ o4 = 2

thus using combinations with repetition we have ( n +r -1 , r)

n=4, r=2

( 4 + 2 - 1 , 2 ) = (5,2) = 10 ways

Comments

i think we can not apply above approach as all employees are distinct for this all items to be distributed should be identical correct if i am wrong

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i think the employees are not labelled employees but under a common category - ie Employees