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A manager wants to assign 20 workers to four distinct construction jobs. These jobs require 6, 4, 3, and 7 workers respectively. In how many different ways can the manager assign the workers?
 

 Mar 27, 2020
 #1
avatar+1970 
0

Take a look at this: https://www.algebra.com/algebra/homework/Permutations/Permutations.faq.question.1114944.html

 

Hope this helped!

 Mar 27, 2020
 #2
avatar+111330 
+1

Choose  any  6  of the 20  for  any  one  of  the  4 jobs =

C(20, 6) * C(4,1)  ways

 

Choose  any  4  of  the remaining 14   for  any  of the 3 remaining jobs

C(14, 4)  * C(3,1)   ways

 

Choose   any  3  of  the remaining  10  for either  of  the  2 remaining jobs

C(10, 3) * C(2,1)   ways

 

[ Note  the remaining 7 workers will automatically be assigned to the remaining job ]

 

So

 

C(20, 6) * C(4,1)  *  C(14, 4)  * C(3,1) *  C(10, 3) * C(2,1)  =

 

111,740,428,800  ways  ...... (yikes !!!!)

 

cool cool cool

 Mar 27, 2020
 #3
avatar+1970 
0

Isn't that TOO big, Chris?!!! 

CalTheGreat  Mar 27, 2020
 #4
avatar+24995 
+1

A manager wants to assign 20 workers to four distinct construction jobs.
These jobs require 6, 4, 3, and 7 workers respectively.
In how many different ways can the manager assign the workers?

 

\(\text{Let Job$_1 = a$} \\ \text{Let Job$_2 = b$} \\ \text{Let Job$_3 = c$} \\ \text{Let Job$_4 = d$} \\ \text{The word is $\mathbf{aaaaaabbbbcccddddddd}$ (The word-length is $20$ ),$\\$how many differnet words we can have ?}\)

 

multinomial coefficient:

\(\begin{array}{|rcll|} \hline \dfrac{20!}{6!\ 4!\ 3!\ 7!} &=& 4655851200 \\ \hline \end{array}\)

 

In 4655851200 different ways can the manager assign the workers.

 

laugh

 Mar 27, 2020

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