A business has 9 applicants for 2 job openings, both as managers. How many ways can the business select 2 new employees out of the 9 candidates?
Since the job positions are exactly the same, the order doesn’t matter. So, to find the number of ways 2 new employees can be selected from a group of 9 candidates, we find the number of permutations: \({}_9P_2 = 72\)
then divide this by the number of arrangements, which is 2! = 2. There are \({}_9C_2 = 72/2 = 36\) different combinations.
Looking at the full calculation:
\[\begin{align*} {}_9C_2 &= \frac{9!}{2!(9-2)!} \\ &= \frac{9!}{2!7!} \\ &= \frac{~~~~9\times 8\times 7\times 6\times 5\times 4\times 3\times 2\times 1}{(2\times 1)\times(7\times 6\times 5\times 4\times 3\times 2\times 1)} \\ &= \frac{9\times 8}{2\times 1} \\ &= \frac{72}{2} \\ &= \mathbf{36} \end{align*}\]