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Five applicants are simultaneously applying for two different jobs in a company. In how many ways can these jobs be filled?
show your solution​


Five Applicants Are Simultaneously Applying For Two Different Jobs In A Company In How Many Ways Can These Jobs Be Filledshow Your Solution class=

Sagot :

[tex]\bold {PROBLEM:}[/tex]

Five applicants are simultaneously applying for two different jobs in a company. In how many ways can these jobs be filled?

[tex]\bold {SOLUTION:}[/tex]

Using the Combinations formula,

[tex] \begin{array}{l} \large \tt C(n,r)= \frac{n!}{(n-r)!r!} \\ \\ \large\tt C(5,2) = \frac{5!}{(5-2)!2!} \\ \\ \large \tt C(5,2)= \frac{5!}{3!2!} \\ \\ \large\tt C(5,2)= \frac{5×4×\cancel{3!}}{\cancel{3!}2!} \\ \\ \large\tt C(5,2) = \frac{20}{2} \\ \\ \large \red{ \boxed{\tt C(5,2)=10}} \end{array}[/tex]

[tex]\bold {FINAL\:ANSWER:}[/tex]

  • There are 10 ways to fill the job.

[tex]\\[/tex]

#CarryOnLearning

[tex]\begin{aligned}\bold{FORMULA:}\\ \\ \sf \: C_k(n)=\bigg(\frac{n}{k}\bigg)=\frac{n!}{k!(n-k)!} \\ \\ \sf \: n = 5 \\ \sf \: k = 2 \\ \\ \sf \: C_2(5)=\bigg(\frac{5}{2}\bigg)=\frac{5!}{2!(5-2)!}\\\\\sf C_2(5)= \frac{5 \times 4}{2 \times 1}= \sf \frac{20}{2} = 10\end{aligned}[/tex]