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A company has 7 job openings and 18 qualified candidates. Each candidate can only be selected for one position. In how many ways can you choose a combination of candidates to fill the job positions?
 

Option: 1

18


Option: 2

35


Option: 3

49


Option: 4

56


Answers (1)

best_answer

Since each candidate can only be selected for one position, we need to calculate the combinations without repetition. We can use the formula for combinations without repetition:

\mathrm{C(n, r)=\frac{n !}{(r ! \times(n-r) !) }}

Here, n is the total number of candidates (18) and r is the number of positions to fill (7).

Using the formula, we have:

\mathrm{C(18,7)=\frac{18 ! }{(7 ! \times(18-7) !)} }

Calculating the factorials:

\mathrm{(18-7) !=11 ! }

Calculating the numerator and denominator:
\mathrm{C(18,7)=\frac{222,768}{5,040} }

Simplifying the fraction:

\mathrm{C(18,7)=35 }

Hence, 35, which represents the number of ways to choose a combination of candidates to fill the 7 job positions while ensuring that each candidate is selected for only one position.

Posted by

Suraj Bhandari

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