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?
+2094 Take a look at this: https://www.algebra.com/algebra/homework/Permutations/Permutations.faq.question.1114944.html
Hope this helped!
+128566 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 !!!!)
+26367 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.
